Dijkstra算法求有向图的最短路径的c语言代码

请看下面的C语言代码实现：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define MAX_NODES 100

int costs[MAX_NODES][MAX_NODES]; // 记录边的权值
int dist[MAX_NODES]; // 记录最短路径长度
int visited[MAX_NODES] = {0}; // 标记此节点是否被访问过
int predecessor[MAX_NODES]; // 记录每个节点的前驱节点

int min_distance(int n) // 从未访问过的节点中找到最短距离的节点
{
    int min_dist = INT_MAX;
    int min_index = -1;
    int i;
    for (i = 0; i < n; i++)
        if (!visited[i] && dist[i] < min_dist)
        {
            min_dist = dist[i];
            min_index = i;
        }
    return min_index;
}

void dijkstra(int start_node, int n)
{
    int i, cur_node;
    for (i = 0; i < n; i++)
    {
        dist[i] = INT_MAX;
        visited[i] = 0;
    }
    dist[start_node] = 0;
    predecessor[start_node] = -1;
    for (i = 0; i < n-1; i++) // 迭代n-1次
    {
        cur_node = min_distance(n); // 获取距离源点最近的未访问过的节点
        visited[cur_node] = 1; // 标记为已访问
        int j;
        for (j = 0; j < n; j++) // 更新与当前节点相邻的节点的距离
            if (!visited[j] && costs[cur_node][j] && dist[cur_node] != INT_MAX &&
                dist[cur_node]+costs[cur_node][j] < dist[j])
            {
                dist[j] = dist[cur_node] + costs[cur_node][j];
                predecessor[j] = cur_node;
            }
    }
}

int main()
{
    int num_of_nodes, start_node, end_node, i, j;

    printf("请输入节点数：");
    scanf("%d", &num_of_nodes);

    printf("请输入节点之间的边以及权值，0代表不连通:\n");
    for (i = 0; i < num_of_nodes; i++)
        for (j = 0; j < num_of_nodes; j++)
            scanf("%d", &costs[i][j]);

    printf("请输入起点：");
    scanf("%d", &start_node);

    printf("请输入终点：");
    scanf("%d", &end_node);

    dijkstra(start_node, num_of_nodes);

    printf("从起点%d到终点%d的最短路径长度为：%d\n", start_node, end_node, dist[end_node]);

    printf("最短路径为：\n");
    int path[MAX_NODES], path_length = 0;
    i = end_node;
    while (i != -1)
    {
        path[path_length++] = i;
        i = predecessor[i];
    }
    for (i = path_length-1; i >= 0; i--)
        printf("%d ", path[i]);

    return 0;
}