Initial commit: C++ courseware

2.Course/2.深入浅出程序设计竞赛（提高篇）/最短路/C++源码/01_dijkstra_naive.cpp

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+// P3371：非负边权的单源最短路，邻接表朴素实现。
+// C++11；时间 O(n*n+m)，空间 O(n+m)。
+#include <iostream>
+#include <vector>
+using namespace std;
+using ll = long long;
+
+struct Edge {
+    int to;
+    ll w;
+};
+
+int main() {
+    ios::sync_with_stdio(false);
+    cin.tie(nullptr);
+    int n, m, s;
+    if (!(cin >> n >> m >> s)) return 0;
+    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);
+    vector<bool> used(n + 1, false);
+    dis[s] = 0;  // 源点可一步不走；其他点尚未找到路线。
+    for (int round = 0; round < n; ++round) {
+        int u = -1;
+        for (int v = 1; v <= n; ++v) {
+            if (!used[v] && (u == -1 || dis[v] < dis[u])) {
+                u = v;
+            }
+        }
+        // 最小候选都不可达时，其余未确定点也不可达。
+        if (u == -1 || dis[u] == INF) break;
+        used[u] = true;
+        for (const Edge& e : g[u]) {
+            ll nd = dis[u] + e.w;
+            if (nd < dis[e.to]) dis[e.to] = nd;
+        }
+    }
+    for (int v = 1; v <= n; ++v) {
+        if (v > 1) cout << ' ';
+        cout << (dis[v] == INF ? 2147483647LL : dis[v]);
+    }
+    cout << '\n';
+    return 0;
+}