A29680.最短距离查询【Floyd】

$时间复杂度：O(n^3)$

#include <bits/stdc++.h>

using namespace std;

int n, m, k, a, b, c, dis[3505][3505],x,y,z;

int main(){
    cin >> n >> m >> k;
    memset(dis,0x3f,sizeof dis);
    
    while(m--){
        cin >> x >> y >> z;
        dis[x][y] = z; 
    }
    
    for(int k = 1;k <= n;k++){
        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= n;j++){
                dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]);
            }
        }
    }
    for(int i = 1;i <= k;i++){
        cin >> a >> b;
        if(dis[a][b] == 0x3f3f3f3f) cout << "impossible" << '\n';
        else cout << dis[a][b] << '\n';
    }
    return 0;
}