题目
floyd算法求所有顶点之间的最短路，典型的模板题。唯一需要注意的是两个顶点之间可能有多条边直接相连，在初始化的时候，直接选择最小的长度作为两点间的距离即可。
实现
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<unordered_map>
#include<list>
#include<string>
#include<string.h>
#include<set>
using namespace std;
int min_dist[105][105]; int main(){
int n, m, u, v, d;
scanf("%d %d", &n, &m);
memset(min_dist, 0x0F, sizeof(min_dist));
for (int i = 1; i <= n; i++)
min_dist[i][i] = 0;
for (int i = 0; i < m; i++){
scanf("%d %d %d", &u, &v, &d);
if (min_dist[u][v] > d) //可能存在两个点之间有多条直接相连的边，取最小的那一条即可
min_dist[u][v] = min_dist[v][u] = d;
}
//floyd算法求最短路
for (int k = 1; k <= n; k++){
for (int i = 1; i <= n; i++){
for (int j = 1; j <= n; j++){
if (min_dist[i][j] > min_dist[i][k] + min_dist[k][j])
min_dist[i][j] = min_dist[i][k] + min_dist[k][j];
}
}
}
for (int i = 1; i <= n; i++){
for (int j = 1; j <= n; j++)
printf("%d ", min_dist[i][j]);
printf("\n");
}
return 0;
}