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;
}