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/07_07_p6464.cpp

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+// 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';
+}