【模板】全源最短路(Johnson All-Pairs Shortest Paths) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<vector<pair<int, long long>>> adj(n + 1);
    for (int i = 0; i < m; ++i) {
        int u, v;
        long long w;
        cin >> u >> v >> w;
        adj[u].push_back({v, w});
    }
    vector<long long> h(n + 1, 0);
    vector<int> in_q(n + 1, 1), cnt(n + 1, 0);
    queue<int> q;
    for (int i = 1; i <= n; ++i) q.push(i);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        in_q[u] = 0;
        for (auto& edge : adj[u]) {
            int v = edge.first;
            long long w = edge.second;
            if (h[u] + w < h[v]) {
                h[v] = h[u] + w;
                cnt[v] = cnt[u] + 1;
                if (cnt[v] >= n) {
                    cout << -1 << "\n";
                    return 0;
                }
                if (!in_q[v]) {
                    in_q[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    const long long INF = 1000000000LL;
    for (int s = 1; s <= n; ++s) {
        vector<long long> d(n + 1, INF);
        priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
        d[s] = 0;
        pq.push({0, s});
        while (!pq.empty()) {
            auto [du, u] = pq.top();
            pq.pop();
            if (du != d[u]) continue;
            for (auto& edge : adj[u]) {
                int v = edge.first;
                long long nw = edge.second + h[u] - h[v];
                if (du + nw < d[v]) {
                    d[v] = du + nw;
                    pq.push({d[v], v});
                }
            }
        }
        long long ans = 0;
        for (int j = 1; j <= n; ++j) {
            long long real_d = (d[j] >= INF) ? INF : (d[j] - h[s] + h[j]);
            ans += 1LL * j * real_d;
        }
        cout << ans << "\n";
    }
    return 0;
}