Codeforces Round #677 (Div. 3) G. Reducing Delivery Cost（dijkstra算法）

题目链接：https://codeforces.com/contest/1433/problem/G

题解

跑 \(n\) 遍 \(dijkstra\) 得到任意两点间的距离，然后枚举哪一条边权为 \(0\) 即可。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m, k;
    cin >> n >> m >> k;
    vector<vector<pair<int, int>>> G(n);
    vector<tuple<int, int, int>> edges(m);
    for (auto &[u, v, w] : edges) {
        cin >> u >> v >> w;
        --u, --v;
        G[u].emplace_back(w, v);
        G[v].emplace_back(w, u);
    }
    vector<vector<int>> dis(n, vector<int> (n));
    auto dijkstra = [&](int s, vector<int> &d) {
        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pque;
        vector<bool> vis(n);
        pque.emplace(0, s);
        while (not pque.empty()) {
            auto [w1, u] = pque.top();
            pque.pop();
            if (vis[u]) continue;
            d[u] = w1;
            vis[u] = true;
            for (auto [w2, v] : G[u]) {
                pque.emplace(w1 + w2, v);
            }
        }
    };
    for (int i = 0; i < n; i++) {
        dijkstra(i, dis[i]);
    }
    int ans = 0;
    vector<pair<int, int>> route(k);
    for (auto &[a, b] : route) {
        cin >> a >> b;
        --a, --b;
        ans += dis[a][b];
    }
    for (auto [u, v, w] : edges) {
        int now = 0;
        for (auto [a, b] : route) {
            now += min({dis[a][b], dis[a][u] + dis[v][b], dis[a][v] + dis[u][b]});
        }
        ans = min(ans, now);
    }
    cout << ans << "\n";
    return 0;
}