有尝求Floyd 算法编程，演算我的手写过程！

Question

急 本人不会编程，是在校高中生。
数学论文用的是弗洛伊德矩阵摹乘法算的做的一个选址问题的最优化。
想用Floyd Warshall algorithm去验算一下算出来的结果是否是正确的，哪位可以有尝帮我编一下这方面的程序吗，原始数据如下。

算法开始：

算法结束数据

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Answer 1

经验证，你手算的结果是正确的。

// C Program for Floyd Warshall Algorithm
#include <stdio.h>

// Number of vertices in the graph
#define V 7

/* Define Infinite as a large enough
value. This value will be used
for vertices not connected to each other */
#define INF 99999

// A function to print the solution matrix
void printSolution(int dist[][V]);

// Solves the all-pairs shortest path
// problem using Floyd Warshall algorithm
void floydWarshall(int graph[][V])
{
    /* dist[][] will be the output matrix
    that will finally have the shortest
    distances between every pair of vertices */
    int dist[V][V], i, j, k;

    /* Initialize the solution matrix
    same as input graph matrix. Or
    we can say the initial values of
    shortest distances are based
    on shortest paths considering no
    intermediate vertex. */
    for (i = 0; i < V; i++)
        for (j = 0; j < V; j++)
            dist[i][j] = graph[i][j];

    /* Add all vertices one by one to
    the set of intermediate vertices.
    ---> Before start of an iteration, we
    have shortest distances between all
    pairs of vertices such that the shortest
    distances consider only the
    vertices in set {0, 1, 2, .. k-1} as
    intermediate vertices.
    ----> After the end of an iteration,
    vertex no. k is added to the set of
    intermediate vertices and the set
    becomes {0, 1, 2, .. k} */
    for (k = 0; k < V; k++)
    {
        // Pick all vertices as source one by one
        for (i = 0; i < V; i++)
        {
            // Pick all vertices as destination for the
            // above picked source
            for (j = 0; j < V; j++)
            {
                // If vertex k is on the shortest path from
                // i to j, then update the value of
                // dist[i][j]
                if (dist[i][k] + dist[k][j] < dist[i][j])
                    dist[i][j] = dist[i][k] + dist[k][j];
            }
        }
    }

    // Print the shortest distance matrix
    printSolution(dist);
}

/* A utility function to print solution */
void printSolution(int dist[][V])
{
    printf(
        "The following matrix shows the shortest distances"
        " between every pair of vertices \n");
    for (int i = 0; i < V; i++)
    {
        for (int j = 0; j < V; j++)
        {
            if (dist[i][j] == INF)
                printf("%7s", "INF");
            else
                printf("%7d", dist[i][j]);
        }
        printf("\n");
    }
}

// driver's code
int main()
{
    int graph[V][V] = {
        {0, INF, INF, INF, 10, INF, 5},
        {INF, 0, 6, 7, 9, INF, INF},
        {INF, 6, 0, 6, 4, INF, INF},
        {INF, 7, 6, 0, INF, 8, 5},
        {10, 9, 4, INF, 0, 7, 7},
        {INF, INF, INF, 8, 7, 0, INF},
        {5, INF, INF, 5, 7, INF, 0}};

    // Function call
    floydWarshall(graph);
    return 0;
}

$ gcc -Wall main.c
$ ./a.out
The following matrix shows the shortest distances between every pair of vertices 
      0     17     14     10     10     17      5
     17      0      6      7      9     15     12
     14      6      0      6      4     11     11
     10      7      6      0     10      8      5
     10      9      4     10      0      7      7
     17     15     11      8      7      0     13
      5     12     11      5      7     13      0