AcWing 854. Floyd求最短路 多源 邻接矩阵

//不存在负权回路
//边权可能为负数
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 210, INF = 1e9;
int n, m, Q;
int d[N][N];//邻接矩阵
void floyd() {//动态规划 
    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() {
    scanf("%d%d%d", &n, &m, &Q);
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            if (i == j) d[i][j] = 0;//处理自环 
            else d[i][j] = INF;
    while (m -- ) {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);
        d[a][b] = min(d[a][b], w);
    }
    floyd();
    while (Q -- ) {
        int a, b;
        scanf("%d%d", &a, &b);
        int t = d[a][b];
        if (t > INF / 2) puts("impossible");
        else printf("%d
", t);
    }
    return 0;
}