Initial commit: C++ courseware
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// P3371：非负边权的单源最短路，邻接表朴素实现。
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// C++11；时间 O(n*n+m)，空间 O(n+m)。
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#include <iostream>
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#include <vector>
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using namespace std;
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using ll = long long;
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struct Edge {
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int to;
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ll w;
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};
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s;
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if (!(cin >> n >> m >> s)) return 0;
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vector<vector<Edge>> g(n + 1);
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for (int i = 0; i < m; ++i) {
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int u, v;
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ll w;
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cin >> u >> v >> w;
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g[u].push_back({v, w});
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}
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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vector<bool> used(n + 1, false);
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dis[s] = 0; // 源点可一步不走；其他点尚未找到路线。
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for (int round = 0; round < n; ++round) {
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int u = -1;
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for (int v = 1; v <= n; ++v) {
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if (!used[v] && (u == -1 || dis[v] < dis[u])) {
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u = v;
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}
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}
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// 最小候选都不可达时，其余未确定点也不可达。
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if (u == -1 || dis[u] == INF) break;
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used[u] = true;
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for (const Edge& e : g[u]) {
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ll nd = dis[u] + e.w;
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if (nd < dis[e.to]) dis[e.to] = nd;
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}
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}
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for (int v = 1; v <= n; ++v) {
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if (v > 1) cout << ' ';
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cout << (dis[v] == INF ? 2147483647LL : dis[v]);
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}
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cout << '\n';
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return 0;
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}
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@@ -0,0 +1,52 @@
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// P3371：相同的朴素 Dijkstra，使用链式前向星保存图。
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// 顶点编号为 1..n；边编号 0 表示边链结束。
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#include <iostream>
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#include <vector>
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using namespace std;
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using ll = long long;
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s;
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if (!(cin >> n >> m >> s)) return 0;
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vector<int> head(n + 1, 0), to(m + 5), nxt(m + 5);
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vector<ll> weight(m + 5);
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int cnt = 0;
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auto add = [&](int u, int v, ll w) {
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++cnt;
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to[cnt] = v;
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weight[cnt] = w;
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nxt[cnt] = head[u]; // 新边先接上旧表头，再更新表头。
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head[u] = cnt;
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};
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for (int i = 0; i < m; ++i) {
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int u, v;
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ll w;
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cin >> u >> v >> w;
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add(u, v, w);
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}
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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vector<bool> used(n + 1, false);
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dis[s] = 0;
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for (int round = 0; round < n; ++round) {
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int u = -1;
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for (int v = 1; v <= n; ++v) {
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if (!used[v] && (u == -1 || dis[v] < dis[u])) u = v;
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}
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if (u == -1 || dis[u] == INF) break;
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used[u] = true;
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for (int i = head[u]; i; i = nxt[i]) {
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int v = to[i];
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ll nd = dis[u] + weight[i];
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if (nd < dis[v]) dis[v] = nd;
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}
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}
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for (int v = 1; v <= n; ++v) {
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if (v > 1) cout << ' ';
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cout << (dis[v] == INF ? 2147483647LL : dis[v]);
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}
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cout << '\n';
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return 0;
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}
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@@ -0,0 +1,63 @@
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// 教学扩展：非负权有向图中，求 s 到 t 的距离并恢复一条最短路线。
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// 输入第一行 n m s t，接着 m 行 u v w。与 P3371 的输入不同。
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// 可达：第一行输出距离，第二行输出路线中的顶点编号。
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// 不可达：只输出 UNREACHABLE。C++11；时间 O(n*n+m)。
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#include <algorithm>
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#include <iostream>
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#include <vector>
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using namespace std;
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using ll = long long;
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struct Edge {
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int to;
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ll w;
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};
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s, t;
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if (!(cin >> n >> m >> s >> t)) return 0;
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vector<vector<Edge>> g(n + 1);
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for (int i = 0; i < m; ++i) {
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int u, v;
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ll w;
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cin >> u >> v >> w;
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g[u].push_back({v, w});
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}
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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vector<bool> used(n + 1, false);
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vector<int> pre(n + 1, -1);
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dis[s] = 0;
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for (int round = 0; round < n; ++round) {
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int u = -1;
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for (int v = 1; v <= n; ++v) {
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if (!used[v] && (u == -1 || dis[v] < dis[u])) u = v;
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}
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if (u == -1 || dis[u] == INF) break;
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used[u] = true;
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for (const Edge& e : g[u]) {
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ll nd = dis[u] + e.w;
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if (nd < dis[e.to]) {
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dis[e.to] = nd;
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pre[e.to] = u; // 与成功改善距离的路线保持一致。
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}
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}
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}
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// 没有路线时不能只把终点放进列表，伪装成一条路径。
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if (dis[t] == INF) {
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cout << "UNREACHABLE\n";
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return 0;
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}
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vector<int> path;
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for (int v = t; v != -1; v = pre[v]) path.push_back(v);
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reverse(path.begin(), path.end());
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cout << dis[t] << '\n';
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for (size_t i = 0; i < path.size(); ++i) {
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if (i) cout << ' ';
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cout << path[i];
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}
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cout << '\n';
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return 0;
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}
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@@ -0,0 +1,46 @@
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// 教材例 10-1 / 洛谷 P3371：非负边权单源最短路。
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// 输入 n m s，随后 m 行 u v w；有向边，可含重边和零权边。
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#include <iostream>
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#include <vector>
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#include <queue>
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#include <utility>
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#include <functional>
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using namespace std;
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using ll = long long;
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using State = pair<ll, int>;
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struct Edge { int to; ll w; };
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s;
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if (!(cin >> n >> m >> s)) return 0;
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vector<vector<Edge>> g(n + 1);
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for (int i = 0; i < m; ++i) {
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int u, v; ll w;
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cin >> u >> v >> w;
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g[u].push_back({v, w});
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}
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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priority_queue<State, vector<State>, greater<State>> q;
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dis[s] = 0;
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q.push({0, s});
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while (!q.empty()) {
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State cur = q.top();
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q.pop();
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ll d = cur.first;
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int u = cur.second;
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if (d != dis[u]) continue; // 旧记录不再扫描出边
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for (const auto &e : g[u]) {
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int v = e.to;
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ll nd = d + e.w;
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if (nd < dis[v]) {
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dis[v] = nd;
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q.push({nd, v});
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}
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}
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}
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for (int i = 1; i <= n; ++i)
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cout << (dis[i] == INF ? 2147483647LL : dis[i])
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<< (i == n ? '\n' : ' ');
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}
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@@ -0,0 +1,34 @@
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// 备份版 Bellman–Ford；输入必须没有源点可达的负环。
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// 输入输出同 P3371；可以用自拟负权无环图验证，未包含判环功能。
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#include <iostream>
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#include <vector>
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using namespace std;
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using ll = long long;
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struct Edge { int u, v; ll w; };
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s;
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if (!(cin >> n >> m >> s)) return 0;
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vector<Edge> edges(m);
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for (auto &e : edges) cin >> e.u >> e.v >> e.w;
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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dis[s] = 0;
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for (int k = 1; k < n; ++k) {
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vector<ll> old = dis;
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bool changed = false;
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for (const auto &e : edges) {
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if (old[e.u] == INF) continue;
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ll nd = old[e.u] + e.w;
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if (nd < dis[e.v]) {
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dis[e.v] = nd;
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changed = true;
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}
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}
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if (!changed) break;
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}
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for (int i = 1; i <= n; ++i)
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cout << (dis[i] == INF ? 2147483647LL : dis[i])
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<< (i == n ? '\n' : ' ');
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}
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@@ -0,0 +1,46 @@
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// FIFO SPFA；输入必须没有源点可达的负环，最坏时间 O(nm)。
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// 输入输出同 P3371；一般负权图应使用第 4 份中的判环版本。
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#include <iostream>
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#include <vector>
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#include <queue>
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using namespace std;
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using ll = long long;
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struct Edge { int to; ll w; };
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int main() {
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ios::sync_with_stdio(false);
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cin.tie(nullptr);
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int n, m, s;
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if (!(cin >> n >> m >> s)) return 0;
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vector<vector<Edge>> g(n + 1);
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for (int i = 0; i < m; ++i) {
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int u, v; ll w;
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cin >> u >> v >> w;
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g[u].push_back({v, w});
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}
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const ll INF = (1LL << 60);
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vector<ll> dis(n + 1, INF);
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vector<char> in_queue(n + 1, false);
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queue<int> q;
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dis[s] = 0;
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q.push(s);
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in_queue[s] = true;
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while (!q.empty()) {
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int u = q.front();
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q.pop();
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in_queue[u] = false; // 离队后仍允许再次排队
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for (const auto &e : g[u]) {
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int v = e.to;
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ll nd = dis[u] + e.w;
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if (nd < dis[v]) {
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dis[v] = nd; // 已在队列中也必须更新距离
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if (!in_queue[v]) {
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q.push(v);
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in_queue[v] = true;
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}
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}
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}
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}
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for (int i = 1; i <= n; ++i)
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cout << (dis[i] == INF ? 2147483647LL : dis[i])
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<< (i == n ? '\n' : ' ');
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}
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@@ -0,0 +1,50 @@
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// P4568 / 教材例 10-2：隐式分层图，最多 k 次免费。
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// n m k；s t；m 行 u v w。城市从 0 编号，无向非负边。
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// 课堂扩展约定：如果没有路线，输出 -1；原题未指定此情况的输出。
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#include <iostream>
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#include <vector>
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#include <queue>
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#include <tuple>
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#include <functional>
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#include <algorithm>
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using namespace std;
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using ll = long long;
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using State = tuple<ll, int, int>;
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struct Edge { int to; ll w; };
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int main() {
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ios::sync_with_stdio(false); cin.tie(nullptr);
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int n, m, k, s, t;
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if (!(cin >> n >> m >> k >> s >> t)) return 0;
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vector<vector<Edge>> g(n);
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for (int i = 0; i < m; ++i) {
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int u, v; ll w; cin >> u >> v >> w;
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g[u].push_back({v, w});
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g[v].push_back({u, w});
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}
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const ll INF = (1LL << 60);
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vector<vector<ll>> dis(k + 1, vector<ll>(n, INF));
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priority_queue<State, vector<State>, greater<State>> q;
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dis[0][s] = 0;
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q.push({0, s, 0});
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while (!q.empty()) {
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State cur = q.top(); q.pop();
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ll d = get<0>(cur);
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int u = get<1>(cur), j = get<2>(cur);
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if (d != dis[j][u]) continue;
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for (const auto &e : g[u]) {
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int v = e.to;
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ll nd = d + e.w;
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if (nd < dis[j][v]) {
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dis[j][v] = nd;
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q.push({nd, v, j});
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}
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if (j < k && d < dis[j + 1][v]) {
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dis[j + 1][v] = d;
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q.push({d, v, j + 1});
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}
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}
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}
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ll ans = INF;
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for (int j = 0; j <= k; ++j) ans = min(ans, dis[j][t]);
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cout << (ans == INF ? -1 : ans) << '\n';
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}
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@@ -0,0 +1,46 @@
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// P1629 / 教材例 10-3：有向正权图，题目保证任意两点互相可达。
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// 输入 n m，随后 m 行 u v w；顶点从 1 编号。
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#include <iostream>
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#include <vector>
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#include <queue>
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#include <utility>
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#include <functional>
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using namespace std;
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using ll = long long;
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using State = pair<ll, int>;
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struct Edge { int to; ll w; };
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vector<ll> dijkstra(const vector<vector<Edge>> &g, int s) {
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const ll INF = (1LL << 60);
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vector<ll> dis(g.size(), INF);
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priority_queue<State, vector<State>, greater<State>> q;
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dis[s] = 0; q.push({0, s});
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while (!q.empty()) {
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State cur = q.top(); q.pop();
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ll d = cur.first; int u = cur.second;
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if (d != dis[u]) continue;
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for (const auto &e : g[u]) {
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ll nd = d + e.w;
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if (nd < dis[e.to]) {
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dis[e.to] = nd;
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q.push({nd, e.to});
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}
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}
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}
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return dis;
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}
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int main() {
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ios::sync_with_stdio(false); cin.tie(nullptr);
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int n, m;
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if (!(cin >> n >> m)) return 0;
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vector<vector<Edge>> g(n + 1), rg(n + 1);
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for (int i = 0; i < m; ++i) {
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int u, v; ll w; cin >> u >> v >> w;
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g[u].push_back({v, w});
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rg[v].push_back({u, w});
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}
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vector<ll> out = dijkstra(g, 1);
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vector<ll> back = dijkstra(rg, 1);
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ll ans = 0;
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for (int i = 2; i <= n; ++i) ans += out[i] + back[i];
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cout << ans << '\n';
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}
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@@ -0,0 +1,46 @@
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// 教学扩展：非负有向图，多源最短路。
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// 输入 n m r；下一行 r 个源点编号；随后 m 行 u v w。
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// 顶点编号 1..n，允许重复源点；输出到各点的最近距离，不可达输出 -1。
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||||
#include <iostream>
|
||||
#include <vector>
|
||||
#include <queue>
|
||||
#include <utility>
|
||||
#include <functional>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
using State = pair<ll, int>;
|
||||
struct Edge { int to; ll w; };
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m, r;
|
||||
if (!(cin >> n >> m >> r)) return 0;
|
||||
vector<int> sources(r);
|
||||
for (int &x : sources) cin >> x;
|
||||
vector<vector<Edge>> g(n + 1);
|
||||
for (int i = 0; i < m; ++i) {
|
||||
int u, v; ll w; cin >> u >> v >> w;
|
||||
g[u].push_back({v, w});
|
||||
}
|
||||
const ll INF = (1LL << 60);
|
||||
vector<ll> dis(n + 1, INF);
|
||||
priority_queue<State, vector<State>, greater<State>> q;
|
||||
for (int x : sources) {
|
||||
if (dis[x] == 0) continue;
|
||||
dis[x] = 0;
|
||||
q.push({0, x});
|
||||
}
|
||||
while (!q.empty()) {
|
||||
State cur = q.top(); q.pop();
|
||||
ll d = cur.first; int u = cur.second;
|
||||
if (d != dis[u]) continue;
|
||||
for (const auto &e : g[u]) {
|
||||
ll nd = d + e.w;
|
||||
if (nd < dis[e.to]) {
|
||||
dis[e.to] = nd;
|
||||
q.push({nd, e.to});
|
||||
}
|
||||
}
|
||||
}
|
||||
for (int i = 1; i <= n; ++i)
|
||||
cout << (dis[i] == INF ? -1 : dis[i]) << (i == n ? '\n' : ' ');
|
||||
}
|
||||
@@ -0,0 +1,53 @@
|
||||
// P3385：检测从 1 可达的负环。C++11。
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
#include <queue>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
struct Edge { int v, w; };
|
||||
|
||||
bool hasNegativeCycle(const vector<vector<Edge>>& g) {
|
||||
int n = static_cast<int>(g.size()) - 1;
|
||||
const ll INF = (1LL << 60);
|
||||
vector<ll> dis(n + 1, INF);
|
||||
vector<int> len(n + 1, 0);
|
||||
vector<bool> inq(n + 1, false);
|
||||
queue<int> q;
|
||||
dis[1] = 0;
|
||||
q.push(1); inq[1] = true;
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop();
|
||||
inq[u] = false;
|
||||
for (auto e : g[u]) {
|
||||
int v = e.v;
|
||||
ll nd = dis[u] + e.w;
|
||||
if (nd >= dis[v]) continue;
|
||||
dis[v] = nd;
|
||||
len[v] = len[u] + 1;
|
||||
if (len[v] >= n) return true;
|
||||
if (!inq[v]) {
|
||||
q.push(v);
|
||||
inq[v] = true;
|
||||
}
|
||||
}
|
||||
}
|
||||
return false;
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int T;
|
||||
if (!(cin >> T)) return 0;
|
||||
while (T--) {
|
||||
int n, m; cin >> n >> m;
|
||||
vector<vector<Edge>> g(n + 1);
|
||||
for (int i = 0; i < m; ++i) {
|
||||
int u, v, w; cin >> u >> v >> w;
|
||||
g[u].push_back({v, w});
|
||||
if (w >= 0) g[v].push_back({u, w});
|
||||
}
|
||||
bool bad = hasNegativeCycle(g);
|
||||
cout << (bad ? "YES" : "NO") << '\n';
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,44 @@
|
||||
// 教学版 Bellman-Ford 判环；输入格式与 P3385 一致。C++11。
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
struct Edge { int u, v, w; };
|
||||
|
||||
bool hasNegativeCycle(int n, const vector<Edge>& edges, int s) {
|
||||
const ll INF = (1LL << 60);
|
||||
vector<ll> dis(n + 1, INF);
|
||||
dis[s] = 0;
|
||||
for (int k = 1; k <= n; ++k) {
|
||||
auto old = dis;
|
||||
bool changed = false;
|
||||
for (auto e : edges) {
|
||||
if (old[e.u] == INF) continue;
|
||||
ll nd = old[e.u] + e.w;
|
||||
if (nd < dis[e.v]) {
|
||||
dis[e.v] = nd;
|
||||
changed = true;
|
||||
}
|
||||
}
|
||||
if (!changed) return false;
|
||||
if (k == n) return true;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int T;
|
||||
if (!(cin >> T)) return 0;
|
||||
while (T--) {
|
||||
int n, m; cin >> n >> m;
|
||||
vector<Edge> edges;
|
||||
for (int i = 0; i < m; ++i) {
|
||||
int u, v, w; cin >> u >> v >> w;
|
||||
edges.push_back({u, v, w});
|
||||
if (w >= 0) edges.push_back({v, u, w});
|
||||
}
|
||||
cout << (hasNegativeCycle(n, edges, 1) ? "YES" : "NO") << '\n';
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,50 @@
|
||||
// P5960：求任意一组差分约束可行解。C++11。
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
#include <queue>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
struct Edge { int v, w; };
|
||||
|
||||
bool solve(const vector<vector<Edge>>& g, vector<ll>& dis) {
|
||||
int n = static_cast<int>(g.size()) - 1;
|
||||
dis.assign(n + 1, 0);
|
||||
vector<int> len(n + 1, 0);
|
||||
vector<bool> inq(n + 1, true);
|
||||
queue<int> q;
|
||||
for (int i = 1; i <= n; ++i) q.push(i);
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop();
|
||||
inq[u] = false;
|
||||
for (auto e : g[u]) {
|
||||
int v = e.v;
|
||||
ll nd = dis[u] + e.w;
|
||||
if (nd >= dis[v]) continue;
|
||||
dis[v] = nd;
|
||||
len[v] = len[u] + 1;
|
||||
if (len[v] >= n) return false;
|
||||
if (!inq[v]) {
|
||||
q.push(v); inq[v] = true;
|
||||
}
|
||||
}
|
||||
}
|
||||
return true;
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, m;
|
||||
if (!(cin >> n >> m)) return 0;
|
||||
vector<vector<Edge>> g(n + 1);
|
||||
for (int i = 0; i < m; ++i) {
|
||||
int a, b, w; cin >> a >> b >> w;
|
||||
g[b].push_back({a, w});
|
||||
}
|
||||
vector<ll> dis;
|
||||
if (!solve(g, dis)) cout << "NO\n";
|
||||
else {
|
||||
for (int i = 1; i <= n; ++i)
|
||||
cout << dis[i] << (i == n ? '\n' : ' ');
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,56 @@
|
||||
// P2865：正权无向图，允许重复点和边的严格次短路。C++11。
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
#include <queue>
|
||||
#include <functional>
|
||||
#include <utility>
|
||||
#include <algorithm>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
struct Edge { int v, w; };
|
||||
|
||||
ll secondShortest(const vector<vector<Edge>>& g) {
|
||||
int n = static_cast<int>(g.size()) - 1;
|
||||
const ll INF = (1LL << 60);
|
||||
vector<ll> d1(n + 1, INF);
|
||||
vector<ll> d2(n + 1, INF);
|
||||
using State = pair<ll, int>;
|
||||
priority_queue<State, vector<State>, greater<State>> q;
|
||||
d1[1] = 0;
|
||||
q.push({0, 1});
|
||||
while (!q.empty()) {
|
||||
State cur = q.top(); q.pop();
|
||||
ll d = cur.first;
|
||||
int u = cur.second;
|
||||
if (d != d1[u] && d != d2[u]) continue;
|
||||
for (auto e : g[u]) {
|
||||
int v = e.v;
|
||||
ll nd = d + e.w;
|
||||
if (nd < d1[v]) {
|
||||
// 新最短留下，旧最短继续竞争第二标签。
|
||||
swap(nd, d1[v]);
|
||||
q.push({d1[v], v});
|
||||
}
|
||||
if (d1[v] < nd && nd < d2[v]) {
|
||||
d2[v] = nd;
|
||||
q.push({d2[v], v});
|
||||
}
|
||||
}
|
||||
}
|
||||
return d2[n];
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, r;
|
||||
if (!(cin >> n >> r)) return 0;
|
||||
vector<vector<Edge>> g(n + 1);
|
||||
for (int i = 0; i < r; ++i) {
|
||||
int u, v, w; cin >> u >> v >> w;
|
||||
g[u].push_back({v, w});
|
||||
g[v].push_back({u, w});
|
||||
}
|
||||
// 原题保证严格次短存在。
|
||||
cout << secondShortest(g) << '\n';
|
||||
}
|
||||
@@ -0,0 +1,29 @@
|
||||
// P2910：完整非负矩阵，按规定顺序访问的最小总危险值。
|
||||
#include <algorithm>
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
|
||||
void floyd(vector<vector<ll>>& d, int n) {
|
||||
for (int k = 1; k <= n; ++k)
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j)
|
||||
d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, m;
|
||||
cin >> n >> m;
|
||||
vector<int> a(m);
|
||||
for (int t = 0; t < m; ++t) cin >> a[t];
|
||||
vector<vector<ll>> d(n + 1, vector<ll>(n + 1));
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j) cin >> d[i][j];
|
||||
floyd(d, n);
|
||||
ll ans = 0;
|
||||
for (int t = 1; t < m; ++t) ans += d[a[t - 1]][a[t]];
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,39 @@
|
||||
// 教学模板：有向图，无负环。输入 n m，随后 m 行 u v w。
|
||||
// 输出 n 行最短距离矩阵，不可达写 INF。这个输出约定为教学自拟。
|
||||
// 建议教学数据 n<=500、|w|<=10^9，有限路长和候选之和远小于 INF。
|
||||
#include <algorithm>
|
||||
#include <iostream>
|
||||
#include <vector>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
const ll INF = 1LL << 60;
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, m;
|
||||
cin >> n >> m;
|
||||
vector<vector<ll>> d(n + 1, vector<ll>(n + 1, INF));
|
||||
for (int i = 1; i <= n; ++i) d[i][i] = 0;
|
||||
for (int t = 0; t < m; ++t) {
|
||||
int u, v;
|
||||
ll w;
|
||||
cin >> u >> v >> w;
|
||||
d[u][v] = min(d[u][v], w);
|
||||
}
|
||||
for (int k = 1; k <= n; ++k)
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j) {
|
||||
if (d[i][k] == INF || d[k][j] == INF) continue;
|
||||
ll nd = d[i][k] + d[k][j];
|
||||
d[i][j] = min(d[i][j], nd);
|
||||
}
|
||||
for (int i = 1; i <= n; ++i) {
|
||||
for (int j = 1; j <= n; ++j) {
|
||||
if (j > 1) cout << ' ';
|
||||
if (d[i][j] == INF) cout << "INF";
|
||||
else cout << d[i][j];
|
||||
}
|
||||
cout << '\n';
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,27 @@
|
||||
// B3611：至少经过一条边的可达性，不预先把对角线设为 true。
|
||||
#include <iostream>
|
||||
using namespace std;
|
||||
bool reach[105][105]{};
|
||||
|
||||
void closure(int n) {
|
||||
for (int k = 1; k <= n; ++k)
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j)
|
||||
reach[i][j] = reach[i][j] || (reach[i][k] && reach[k][j]);
|
||||
}
|
||||
|
||||
int main() {
|
||||
ios::sync_with_stdio(false);
|
||||
cin.tie(nullptr);
|
||||
int n, x;
|
||||
cin >> n;
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j) {
|
||||
cin >> x;
|
||||
reach[i][j] = (x != 0);
|
||||
}
|
||||
closure(n);
|
||||
for (int i = 1; i <= n; ++i)
|
||||
for (int j = 1; j <= n; ++j)
|
||||
cout << int(reach[i][j]) << (j == n ? '\n' : ' ');
|
||||
}
|
||||
@@ -0,0 +1,29 @@
|
||||
// P1144 最短路计数。C++17；无向无权图，重边按不同边计数。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
vector<int> head(n + 1, -1), to(2 * m), nxt(2 * m);
|
||||
int tot = 0;
|
||||
auto add = [&](int u, int v) {
|
||||
to[tot] = v; nxt[tot] = head[u]; head[u] = tot++;
|
||||
};
|
||||
for (int i = 0, u, v; i < m; ++i) {
|
||||
cin >> u >> v; add(u, v); add(v, u);
|
||||
}
|
||||
vector<int> dist(n + 1, -1), ways(n + 1, 0);
|
||||
queue<int> q; dist[1] = 0; ways[1] = 1; q.push(1);
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop();
|
||||
for (int e = head[u]; e != -1; e = nxt[e]) {
|
||||
int v = to[e];
|
||||
if (dist[v] == -1) {
|
||||
dist[v] = dist[u] + 1; q.push(v);
|
||||
}
|
||||
if (dist[v] == dist[u] + 1)
|
||||
ways[v] = (ways[v] + ways[u]) % 100003;
|
||||
}
|
||||
}
|
||||
for (int i = 1; i <= n; ++i) cout << ways[i] << '\n';
|
||||
}
|
||||
@@ -0,0 +1,44 @@
|
||||
// P1875 魔法药水。每个原料位置独立选择制法，交换配方书写顺序不重复计数。
|
||||
// 当前原题提示同一对原料可能有不同产物，必须保存所有不同配方。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
ll addSat(ll a, ll b) { return a > LLONG_MAX - b ? LLONG_MAX : a + b; }
|
||||
ll mulSat(ll a, ll b) {
|
||||
if (a == 0 || b == 0) return 0;
|
||||
return a > LLONG_MAX / b ? LLONG_MAX : a * b;
|
||||
}
|
||||
struct Recipe { int a, b, c; };
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n; if (!(cin >> n)) return 0;
|
||||
vector<ll> d(n), ways(n, 1);
|
||||
for (ll &x : d) cin >> x;
|
||||
vector<Recipe> rec;
|
||||
vector<vector<int>> incident(n);
|
||||
unordered_set<ll> seen;
|
||||
int a, b, c;
|
||||
while (cin >> a >> b >> c) {
|
||||
if (a > b) swap(a, b);
|
||||
ll key = (1LL * a * n + b) * n + c;
|
||||
if (!seen.insert(key).second) continue;
|
||||
int id = (int)rec.size(); rec.push_back({a, b, c});
|
||||
incident[a].push_back(id); incident[b].push_back(id);
|
||||
}
|
||||
vector<int> need(rec.size(), 2);
|
||||
vector<bool> done(n, false);
|
||||
for (int round = 0; round < n; ++round) {
|
||||
int u = -1;
|
||||
for (int i = 0; i < n; ++i)
|
||||
if (!done[i] && (u == -1 || d[i] < d[u])) u = i;
|
||||
done[u] = true;
|
||||
for (int id : incident[u]) {
|
||||
if (--need[id] != 0) continue;
|
||||
auto [a, b, c] = rec[id];
|
||||
ll nd = d[a] + d[b], cnt = mulSat(ways[a], ways[b]);
|
||||
if (nd < d[c]) { d[c] = nd; ways[c] = cnt; }
|
||||
else if (nd == d[c]) ways[c] = addSat(ways[c], cnt);
|
||||
}
|
||||
}
|
||||
cout << d[0] << ' ' << ways[0] << '\n';
|
||||
}
|
||||
@@ -0,0 +1,42 @@
|
||||
// P1462 通往奥格瑞玛的道路。城市费二分，非负道路耗血用 Dijkstra。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; ll b; if (!(cin >> n >> m >> b)) return 0;
|
||||
vector<int> fee(n + 1), values;
|
||||
for (int i = 1; i <= n; ++i) { cin >> fee[i]; values.push_back(fee[i]); }
|
||||
vector<vector<pair<int,ll>>> g(n + 1);
|
||||
for (int i = 0, u, v; i < m; ++i) {
|
||||
ll w; cin >> u >> v >> w;
|
||||
g[u].push_back({v,w}); g[v].push_back({u,w});
|
||||
}
|
||||
auto check = [&](int X) {
|
||||
if (fee[1] > X || fee[n] > X) return false;
|
||||
const ll INF = 1LL << 60;
|
||||
vector<ll> dist(n + 1, INF);
|
||||
priority_queue<pair<ll,int>, vector<pair<ll,int>>, greater<pair<ll,int>>> pq;
|
||||
dist[1] = 0; pq.push({0,1});
|
||||
while (!pq.empty()) {
|
||||
auto [du,u] = pq.top(); pq.pop();
|
||||
if (du != dist[u]) continue;
|
||||
for (auto [v,w] : g[u]) {
|
||||
if (fee[v] > X) continue;
|
||||
if (du + w < dist[v]) {
|
||||
dist[v] = du + w; pq.push({dist[v],v});
|
||||
}
|
||||
}
|
||||
}
|
||||
return dist[n] <= b;
|
||||
};
|
||||
sort(values.begin(), values.end());
|
||||
values.erase(unique(values.begin(), values.end()), values.end());
|
||||
if (!check(values.back())) { cout << "AFK\n"; return 0; }
|
||||
int l = 0, r = (int)values.size() - 1;
|
||||
while (l < r) {
|
||||
int mid = l + (r-l)/2;
|
||||
if (check(values[mid])) r = mid; else l = mid + 1;
|
||||
}
|
||||
cout << values[l] << '\n';
|
||||
}
|
||||
@@ -0,0 +1,42 @@
|
||||
// P1073 最优贸易。价格是 1..100，单调标签至多改善 100 次。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
vector<int> price(n + 1);
|
||||
for (int i = 1; i <= n; ++i) cin >> price[i];
|
||||
vector<vector<int>> g(n + 1), rg(n + 1);
|
||||
auto add = [&](int u, int v) { g[u].push_back(v); rg[v].push_back(u); };
|
||||
for (int i = 0, u, v, z; i < m; ++i) {
|
||||
cin >> u >> v >> z; add(u,v); if (z == 2) add(v,u);
|
||||
}
|
||||
vector<int> low(n + 1, 101), high(n + 1, -1);
|
||||
vector<bool> inq(n + 1, false);
|
||||
queue<int> q; low[1] = price[1]; q.push(1); inq[1] = true;
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop(); inq[u] = false;
|
||||
for (int v : g[u]) {
|
||||
int cand = min(low[u], price[v]);
|
||||
if (cand < low[v]) {
|
||||
low[v] = cand;
|
||||
if (!inq[v]) { q.push(v); inq[v] = true; }
|
||||
}
|
||||
}
|
||||
}
|
||||
high[n] = price[n]; q.push(n); inq[n] = true;
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop(); inq[u] = false;
|
||||
for (int v : rg[u]) {
|
||||
int cand = max(high[u], price[v]);
|
||||
if (cand > high[v]) {
|
||||
high[v] = cand;
|
||||
if (!inq[v]) { q.push(v); inq[v] = true; }
|
||||
}
|
||||
}
|
||||
}
|
||||
int ans = 0;
|
||||
for (int v = 1; v <= n; ++v)
|
||||
if (low[v] <= 100 && high[v] >= 1) ans = max(ans, high[v]-low[v]);
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,29 @@
|
||||
// P1993 小K的农场。上界边 x[v] <= x[u]+w，全图判负环。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
vector<vector<pair<int,ll>>> g(n + 1);
|
||||
auto add = [&](int u, int v, ll w) { g[u].push_back({v,w}); };
|
||||
for (int i = 0, type, a, b; i < m; ++i) {
|
||||
cin >> type >> a >> b; ll c;
|
||||
if (type == 1) { cin >> c; add(a,b,-c); }
|
||||
else if (type == 2) { cin >> c; add(b,a,c); }
|
||||
else { add(a,b,0); add(b,a,0); }
|
||||
}
|
||||
vector<ll> dist(n + 1, 0);
|
||||
vector<int> len(n + 1, 0);
|
||||
vector<bool> inq(n + 1, true);
|
||||
queue<int> q; for (int i = 1; i <= n; ++i) q.push(i);
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop(); inq[u] = false;
|
||||
for (auto [v,w] : g[u]) if (dist[v] > dist[u] + w) {
|
||||
dist[v] = dist[u] + w; len[v] = len[u] + 1;
|
||||
if (len[v] >= n) { cout << "No\n"; return 0; }
|
||||
if (!inq[v]) { q.push(v); inq[v] = true; }
|
||||
}
|
||||
}
|
||||
cout << "Yes\n";
|
||||
}
|
||||
@@ -0,0 +1,68 @@
|
||||
// P3275 糖果。迭代 Kosaraju + 缩点 DAG 上的最长下界，O(n+k)。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, k; if (!(cin >> n >> k)) return 0;
|
||||
vector<vector<pair<int,int>>> g(n);
|
||||
vector<vector<int>> rg(n);
|
||||
auto add = [&](int u, int v, int w) { g[u].push_back({v,w}); rg[v].push_back(u); };
|
||||
for (int i = 0, x, a, b; i < k; ++i) {
|
||||
cin >> x >> a >> b; --a; --b;
|
||||
if (x == 1) { add(a,b,0); add(b,a,0); }
|
||||
if (x == 2) add(a,b,1);
|
||||
if (x == 3) add(b,a,0);
|
||||
if (x == 4) add(b,a,1);
|
||||
if (x == 5) add(a,b,0);
|
||||
}
|
||||
vector<bool> vis(n, false);
|
||||
vector<int> order;
|
||||
// 第一次 DFS：显式保存下一条边下标，退出时记录结束序。
|
||||
for (int s = 0; s < n; ++s) if (!vis[s]) {
|
||||
vector<pair<int,int>> st;
|
||||
vis[s] = true; st.push_back({s,0});
|
||||
while (!st.empty()) {
|
||||
auto &[u,i] = st.back();
|
||||
if (i == (int)g[u].size()) { order.push_back(u); st.pop_back(); }
|
||||
else {
|
||||
int v = g[u][i++].first;
|
||||
if (!vis[v]) { vis[v] = true; st.push_back({v,0}); }
|
||||
}
|
||||
}
|
||||
}
|
||||
// 第二次 DFS：反图按逆结束序分组。
|
||||
vector<int> comp(n, -1), size;
|
||||
int count = 0;
|
||||
reverse(order.begin(), order.end());
|
||||
for (int s : order) if (comp[s] == -1) {
|
||||
vector<int> st(1,s); comp[s] = count; int amount = 0;
|
||||
while (!st.empty()) {
|
||||
int u = st.back(); st.pop_back(); ++amount;
|
||||
for (int v : rg[u]) if (comp[v] == -1) {
|
||||
comp[v] = count; st.push_back(v);
|
||||
}
|
||||
}
|
||||
size.push_back(amount); ++count;
|
||||
}
|
||||
vector<vector<pair<int,int>>> dag(count);
|
||||
vector<int> indeg(count, 0), d(count, 1);
|
||||
for (int u = 0; u < n; ++u) for (auto [v,w] : g[u]) {
|
||||
if (comp[u] == comp[v]) {
|
||||
if (w == 1) { cout << -1 << '\n'; return 0; }
|
||||
} else {
|
||||
dag[comp[u]].push_back({comp[v],w}); ++indeg[comp[v]];
|
||||
}
|
||||
}
|
||||
queue<int> q;
|
||||
for (int c = 0; c < count; ++c) if (indeg[c] == 0) q.push(c);
|
||||
while (!q.empty()) {
|
||||
int u = q.front(); q.pop();
|
||||
for (auto [v,w] : dag[u]) {
|
||||
d[v] = max(d[v], d[u]+w);
|
||||
if (--indeg[v] == 0) q.push(v);
|
||||
}
|
||||
}
|
||||
long long ans = 0;
|
||||
for (int c = 0; c < count; ++c) ans += 1LL * size[c] * d[c];
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,29 @@
|
||||
// P6464 传送门。原图 Floyd 一次，独立枚举一对零边。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
const ll INF = 1LL << 60;
|
||||
vector<vector<ll>> d(n, vector<ll>(n, INF));
|
||||
for (int i = 0; i < n; ++i) d[i][i] = 0;
|
||||
for (int e = 0, u, v; e < m; ++e) {
|
||||
ll w; cin >> u >> v >> w; --u; --v;
|
||||
d[u][v] = d[v][u] = min(d[u][v], w);
|
||||
}
|
||||
for (int k = 0; k < n; ++k)
|
||||
for (int i = 0; i < n; ++i)
|
||||
for (int j = 0; j < n; ++j)
|
||||
if (d[i][k] != INF && d[k][j] != INF)
|
||||
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
|
||||
ll ans = 0;
|
||||
for (int i = 0; i < n; ++i) for (int j = i+1; j < n; ++j) ans += d[i][j];
|
||||
for (int u = 0; u < n; ++u) for (int v = u+1; v < n; ++v) {
|
||||
ll sum = 0;
|
||||
for (int i = 0; i < n; ++i) for (int j = i+1; j < n; ++j)
|
||||
sum += min({d[i][j], d[i][u]+d[v][j], d[i][v]+d[u][j]});
|
||||
ans = min(ans, sum);
|
||||
}
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,28 @@
|
||||
// P3403 跳楼机。楼层减一，按最小步长的余数建图。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
ll h; array<int,3> step;
|
||||
if (!(cin >> h >> step[0] >> step[1] >> step[2])) return 0;
|
||||
sort(step.begin(), step.end()); int m = step[0];
|
||||
const ll INF = 1LL << 60;
|
||||
vector<ll> d(m, INF);
|
||||
priority_queue<pair<ll,int>, vector<pair<ll,int>>, greater<pair<ll,int>>> pq;
|
||||
d[0] = 0; pq.push({0,0});
|
||||
while (!pq.empty()) {
|
||||
auto [du,u] = pq.top(); pq.pop();
|
||||
if (du != d[u]) continue;
|
||||
for (int i = 1; i <= 2; ++i) {
|
||||
int w = step[i], v = (u+w)%m;
|
||||
if (du + w < d[v]) { d[v] = du+w; pq.push({d[v],v}); }
|
||||
}
|
||||
}
|
||||
ll L = h-1, ans = 0;
|
||||
for (int r = 0; r < m; ++r) {
|
||||
if (d[r] == INF || d[r] > L) continue;
|
||||
ans += (L-d[r])/m + 1;
|
||||
}
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,28 @@
|
||||
// P1119 灾后重建。原题编号 0..n-1，重建时间和询问时间分别不下降。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
vector<int> repair(n); for (int &t : repair) cin >> t;
|
||||
const ll INF = 1LL << 60;
|
||||
vector<vector<ll>> d(n, vector<ll>(n, INF));
|
||||
for (int i = 0; i < n; ++i) d[i][i] = 0;
|
||||
for (int e = 0, u, v; e < m; ++e) {
|
||||
ll w; cin >> u >> v >> w; d[u][v] = d[v][u] = min(d[u][v],w);
|
||||
}
|
||||
int Q, k = 0; cin >> Q;
|
||||
while (Q--) {
|
||||
int x, y, t; cin >> x >> y >> t;
|
||||
while (k < n && repair[k] <= t) {
|
||||
for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) {
|
||||
if (d[i][k] == INF || d[k][j] == INF) continue;
|
||||
d[i][j] = min(d[i][j], d[i][k]+d[k][j]);
|
||||
}
|
||||
++k;
|
||||
}
|
||||
if (repair[x] > t || repair[y] > t || d[x][y] == INF) cout << -1 << '\n';
|
||||
else cout << d[x][y] << '\n';
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,28 @@
|
||||
// P1037 产生数。十个数字的自反传递闭包 + 十进制高精度乘法。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
string mulSmall(string s, int k) {
|
||||
int carry = 0;
|
||||
for (int i = (int)s.size()-1; i >= 0; --i) {
|
||||
int t = (s[i]-'0')*k + carry;
|
||||
s[i] = char('0'+t%10); carry = t/10;
|
||||
}
|
||||
while (carry) { s.insert(s.begin(), char('0'+carry%10)); carry /= 10; }
|
||||
return s;
|
||||
}
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
string number; int k; if (!(cin >> number >> k)) return 0;
|
||||
bool reach[10][10] = {};
|
||||
for (int i = 0; i < 10; ++i) reach[i][i] = true;
|
||||
for (int i = 0, a, b; i < k; ++i) { cin >> a >> b; reach[a][b] = true; }
|
||||
for (int mid = 0; mid < 10; ++mid)
|
||||
for (int i = 0; i < 10; ++i)
|
||||
for (int j = 0; j < 10; ++j)
|
||||
reach[i][j] |= reach[i][mid] && reach[mid][j];
|
||||
int choices[10] = {};
|
||||
for (int a = 0; a < 10; ++a) for (int b = 0; b < 10; ++b) choices[a] += reach[a][b];
|
||||
string ans = "1";
|
||||
for (char digit : number) ans = mulSmall(ans, choices[digit-'0']);
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,21 @@
|
||||
// P2419 牛大赛。胜者指向败者，闭包后判断是否与其余点全部可比较。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
vector<vector<bool>> reach(n, vector<bool>(n, false));
|
||||
for (int i = 0, a, b; i < m; ++i) { cin >> a >> b; reach[a-1][b-1] = true; }
|
||||
for (int k = 0; k < n; ++k) for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j)
|
||||
reach[i][j] = reach[i][j] || (reach[i][k] && reach[k][j]);
|
||||
int ans = 0;
|
||||
for (int v = 0; v < n; ++v) {
|
||||
int known = 0;
|
||||
for (int j = 0; j < n; ++j) {
|
||||
if (j == v) continue;
|
||||
if (reach[v][j] || reach[j][v]) ++known;
|
||||
}
|
||||
if (known == n-1) ++ans;
|
||||
}
|
||||
cout << ans << '\n';
|
||||
}
|
||||
@@ -0,0 +1,34 @@
|
||||
// P2047 社交网络。正权、无向、简单、连通图；有序点对，排除端点。
|
||||
#include <bits/stdc++.h>
|
||||
using namespace std;
|
||||
using ll = long long;
|
||||
int main() {
|
||||
ios::sync_with_stdio(false); cin.tie(nullptr);
|
||||
int n, m; if (!(cin >> n >> m)) return 0;
|
||||
const ll INF = 1LL << 60;
|
||||
vector<vector<ll>> d(n, vector<ll>(n, INF));
|
||||
// 中间受限图的计数可能超过最终答案范围，从存储起即用浮点。
|
||||
vector<vector<long double>> cnt(n, vector<long double>(n, 0));
|
||||
for (int i = 0; i < n; ++i) d[i][i] = 0;
|
||||
for (int e = 0, u, v; e < m; ++e) {
|
||||
ll w; cin >> u >> v >> w; --u; --v;
|
||||
d[u][v] = d[v][u] = w; cnt[u][v] = cnt[v][u] = 1;
|
||||
}
|
||||
for (int k = 0; k < n; ++k) for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) {
|
||||
if (i == k || j == k || i == j) continue;
|
||||
if (d[i][k] == INF || d[k][j] == INF) continue;
|
||||
ll nd = d[i][k]+d[k][j];
|
||||
long double ways = cnt[i][k]*cnt[k][j];
|
||||
if (nd < d[i][j]) { d[i][j] = nd; cnt[i][j] = ways; }
|
||||
else if (nd == d[i][j]) cnt[i][j] += ways;
|
||||
}
|
||||
cout << fixed << setprecision(3);
|
||||
for (int v = 0; v < n; ++v) {
|
||||
long double ans = 0;
|
||||
for (int s = 0; s < n; ++s) for (int t = 0; t < n; ++t) {
|
||||
if (s == t || s == v || t == v) continue;
|
||||
if (d[s][v]+d[v][t] == d[s][t]) ans += cnt[s][v]*cnt[v][t]/cnt[s][t];
|
||||
}
|
||||
cout << ans << '\n';
|
||||
}
|
||||
}
|
||||
@@ -0,0 +1,51 @@
|
||||
# 源码说明
|
||||
|
||||
共 28 份独立完整程序。建议全部按 GCC C++17 编译。每个文件含独立 `main`，应单独编译。输入来自标准输入，结果写到标准输出。
|
||||
|
||||
## 前六份课件
|
||||
|
||||
| 文件 | 对应内容 | 输入与输出约定 |
|
||||
| --- | --- | --- |
|
||||
| 01_dijkstra_naive.cpp | P3371，邻接表朴素 Dijkstra | n m s；m 条有向边；输出源点到各点距离 |
|
||||
| 01b_dijkstra_forward_star.cpp | P3371，前向星朴素 Dijkstra | 同上；改变存图方式 |
|
||||
| 01c_dijkstra_path.cpp | 第一章路径恢复扩展 | 教学输入格式与路径输出见源码注释及第一章备注 |
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| 02_dijkstra_heap.cpp | P3371，堆优化 Dijkstra | 同上；非负边权 |
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| 02b_bellman_ford.cpp | 备份版 Bellman–Ford | 同 P3371 输入输出；教学扩展允许负边，无源点可达负环 |
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| 02c_spfa.cpp | FIFO SPFA | 同上；此版不带判环，最坏 O(nm) |
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| 03_flights.cpp | P4568，分层状态 | n m k；s t；m 条无向边；0 开始编号；不可达输出 -1 是教学扩展约定 |
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| 03b_postman.cpp | P1629，反图与往返 | n m；m 条有向边；1 开始编号；按原题保证各点互达 |
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| 03c_multi_source.cpp | 多源最短路教学扩展 | n m r；r 个源点；m 条有向边；不可达输出 -1 |
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| 04_negative_cycle.cpp | P3385，SPFA 判环 | 多组数据；仅检测从 1 可达的负环；非负输入边双向、负边单向 |
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| 04a_bf_negative.cpp | P3385，Bellman–Ford 判环 | 同上一文件 |
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| 04b_difference_constraints.cpp | P5960，差分约束 | a b c 表示 x[a]−x[b]≤c；输出任意解，无解为 NO |
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| 05_second_shortest.cpp | P2865，严格次短路 | 正权无向图；允许重复经过点边，第二距离严格大于最短距离 |
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| 06_floyd_trip.cpp | P2910，规定访问顺序 | 读入访问序列和完整危险值矩阵，输出最小总危险值 |
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| 06a_floyd_general.cpp | 通用 Floyd 教学模板 | n m；m 条有向边；无负环；输出矩阵，不可达为 INF |
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| 06b_transitive_closure.cpp | B3611，非空路径闭包 | 输入邻接矩阵；不预先把对角线设为真 |
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P3371 系列程序以 2147483647 表示不可达。通用 Floyd 教学模板建议 n≤500、边权绝对值≤10^9，确保所选 INF 和 `long long` 有足够余量。
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## 综合习题的重要前提
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`07_01_p1144.cpp` 至 `07_12_p2047.cpp` 与课程使用说明的 12 道习题一一对应，输入输出遵循相应题号。
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- P1144：重边按不同边计数，模 100003。
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- P1875：配方读到 EOF；保存同一原料对的不同产物，完全重复的无序配方去重。主算法 O(n²+R)，哈希去重的 O(R) 是期望复杂度。无关药水极大中间计数采用饱和运算，原题承诺范围内的目标答案保持精确。
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- P1462：费用包括起终点，耗血恰好等于初始血量仍可行。
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- P1073：传播次数的界依赖价格范围 1…100。
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- P1993：全图判定约束可行性，所有点一起初始化。
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- P3275：非递归强连通分量与缩点，避免深链递归栈溢出；总量用 `long long`。
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- P6464：每次试装传送门都读取原图 Floyd 距离。
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- P3403：从第 1 层开始计数，先排除不可达再计数，覆盖 h=2^63−1 边界。
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- P1119：依赖原题重建时间、询问时间分别不降的保证；0 开始编号。
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- P1037：允许零次变换，对角线置真；按最终数字去重，使用十进制高精度字符串。
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||||
- P2419：原题胜负关系无矛盾，检查是否与其余所有点可比较。
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- P2047：正权简单无向图；有序点对双向累计并排除端点，使用 `long double` 计数与求比例，输出三位小数。
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## 验证
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|
||||
本次以 C++17 重新编译第 2—7 章的 25 份程序，31 组新增教学输入及关键边界检查全部通过。输入输出见 `../教学样例/样例清单.json`，对应记录见 `../资料与核查/修订样例验证.json`。
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第一章 3 份程序的修订验证记录另见 `../资料与核查/第一章程序核查.json`。旧版 27 份程序累计 3,972 组本地对照记录保留在 `../资料与核查/旧版程序验证`，本次没有重复执行全部历史测试。
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所有验证均为本地编译运行，没有向在线评测提交。
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