floyd算法用来求多源汇最短路

用邻接矩阵来存所有的边

时间复杂度O(n^3)

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

using namespace std;

const int N = 20010,INF = 1e9;

int n,m,k;
int g[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 ++ ){
                g[i][j] = min(g[i][j],g[i][k] + g[k][j]);
            }
        }
    }
}

int main(){
    cin >> n >> m >> k;
    for(int i = 1;i <= n;i ++ ){
        for(int j = 1;j <= m;j ++ ){
            if(i == j) g[i][j] = 0;
            else g[i][j] = INF;
        }
    }
    for(int i = 0;i < m;i ++ ){
        int x,y,z;
        cin >> x >> y >> z;
        g[x][y] = min(g[x][y],z);
    }
    floyd();
    while(k -- ){
        int x,y;
        cin >> x >> y;
        if(g[x][y] > INF / 2) cout << "impossible" << endl; //INF还是要/2，考虑到可能有负权边的情况
        else cout << g[x][y] <<endl;
    }
    return 0;
}