Floyd算法求多流最短路径

Floyd算法求多源最短路径

#include<stdio.h>

int main()
{
	int m, n, i, j, k, t1, t2, t3, e[10][10];
	scanf_s("%d %d", &n, &m);
	for (i = 1; i <= n;i++)
	for (j = 1; j <= n; j++)//Initialize the matrix
	{
		if (i == j)
			e[i][j] = 0;
		else
			e[i][j] = 99999;
	}
	for (i = 1; i <= m; i++)//read the vertex
	{
		scanf_s("%d %d %d", &t1, &t2, &t3);
		e[t1][t2] = t3;
	}

	for (k = 1; k <= n;k++)
	for (i = 1; i <= n;i++)
	for (j = 1; j <= n; j++)
	{
		if (e[i][j] > e[i][k] + e[k][j])
			e[i][j] = e[i][k] + e[k][j];
	}

	for (i = 1; i <= n; i++)
	{
		for (j = 1; j <= n; j++)
			printf_s("%d ", e[i][j]);
		printf_s("\n");
	}

	getchar();
	return 0;
}

    n代表节点数目，m代表边的数目。输入的时候t1 t2代表两个边，t3代表权值，k代表中间节点，最后就可以输出任意两点之间的最短路径(代码来源于纪磊)。