题目链接：Floyd求最短路

#include <iostream>
#include <algorithm>
#include <cstring>

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;
    
    for(int i = 0; i < m; i++)
    {
        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) cout << "impossible" << endl;
        else cout << d[a][b] << endl;
    }
    return 0;
}