Floyd算法解决最短路径问题

本文给出了用Floyd算法来求最短路径问题的程序。

输入如下所示：

1 —————–测试用例个数
10 —————–本测试用例的点数
1 2 4 ——————第一个点和第二个点之间的距离是4.
1 3 8
2 3 3
2 4 4
2 5 6
3 4 2
3 5 2
4 5 4
4 6 9
5 6 4
代码如下：

#include 
#define MAX 6
#define MAX_VALUE 65536

int v[MAX + 1][MAX + 1] = { 0 };
int p[MAX + 1][MAX + 1] = { 0 };
void Floyd()
{
    v[1][1] = 0;
    for (int k = 1; k <= MAX; k++)
    for (int i = 1; i <= MAX; i++)
    for (int j = 1; j <= MAX; j++)
        if (v[i][k] + v[k][j]for (int i = 1; i <= MAX; i++)
        for (int j = 1; j <= MAX; j++)
        {
            v[i][j] = MAX_VALUE;
        }
}
void printPathFloyd()
{
    printf("%d\n", v[1][6]);
    int path[10];
    int end = 0;
    int curValue = 6;
    while (curValue != 0)
    {
        path[end++] = curValue;
        curValue = p[1][curValue];
    }
    for (int j = end - 1; j >= 0; j--)
    {
        printf("%d ", path[j]);
    }
    printf("\n");
}

int main(int argc, char** argv)
{
    freopen("input.txt", "r", stdin);
    int N;
    scanf("%d\n", &N);

    for (int case_num = 0; case_num < N; case_num++)
    {
        initV();
        int line_num;
        scanf("%d\n", &line_num);
        for (int i = 0; i < line_num; i++)
        {
            int pt1, pt2, length;
            scanf("%d %d %d", &pt1, &pt2, &length);
            v[pt1][pt2] = length;
            v[pt2][pt1] = length;
        }
        Floyd();
        printPathFloyd();
    }
}