AcWing854.Floyd求最短路

題解

注意:k, i, j的順序不能改變，必須以k作為橋樑

#include <iostream>
#include <cstring>
using namespace std;

const int N = 210;

int g[N][N], n, m, k;

int main()
{
    cin >> n >> m >> k;
    
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
            if(i == j) g[i][j] = 0; //刪去自環，不讓自環成為一條大於0的路徑
            else g[i][j] = 0x3f3f3f3f;
    
    int a, b, c;
    for(int i = 0; i < m; ++i)
    {
        cin >> a >> b >> c;
        g[a][b] = min(g[a][b], c);  //因為有重邊
    }
    for(int k = 1; k <= n; ++k)
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
    for(int i = 0; i < k; ++i)
    {
        cin >> a >> b;
        if(g[a][b] >= 0x3f3f3f3f/2) cout << "impossible" << endl;
        else cout << g[a][b] << endl;
    }
    return 0;
}