854 floyd路径

#include <bits/stdc++.h>
using namespace std;

const int N = 210, INF = 1e9;

int n, m, Q;
int d[N][N];

void floyd() {
    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() {
    cin >> n >> m >> Q;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == j) d[i][j] = 0;
            else d[i][j] = INF;
        }
    }
    
    while(m --) {
        int a, b, w;
        cin >> a >> b >> w;
        d[a][b] = min(d[a][b], w);
    }
    floyd();
    
    while(Q --) {
        int a, b;
        cin >> a >> b;
        if (d[a][b] > INF / 2) puts("impossible");
        else cout << d[a][b] << endl;
    }
    return 0;
}
posted @ 2022-11-11 11:26  天然气之子  阅读(13)  评论(0编辑  收藏  举报