hiho_1089_floyd最短路

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

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#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; }