Floy

多元最短路——Floyd

思路：枚举起点终点和中转点

变量：

int m, dis[N][N], n;

m条边，n个点，dis[i][j]数组储存从i到j的最短路径

主体：

void Folyd() {
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (dis[i][j] > dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
    return;
}

分解一下。。。

        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)

for循环枚举从i到j的路径

枚举中转点

for (int k = 1; k <= n; k++)

完整代码：

#include <iostream>
#include <cstdio>
#define N 1001
#define INF 233

int m, dis[N][N], n;

void Folyd() {
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (dis[i][j] > dis[i][k] + dis[k][j])
                    dis[i][j] = dis[i][k] + dis[k][j];
    return;
}

int main()
{
    scanf("%d %d", &n, &m);    
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            dis[i][j] = INF;
            if (i == j)
                dis[i][j] = 0;
        } 
    for (int i = 1; i <= m; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        dis[a][b] = c;
    }
    Folyd();
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            printf("%5d", dis[i][j]);
        }
        printf("\n\n");
    }

    return 0;
}