c语言中根据kruskal算法，实现求最小生成树代码.并设计案例调用

以下是一个简单的案例，使用Kruskal算法求解一个带权无向图的最小生成树：

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

#define MAX_VERTICES 100
#define MAX_EDGES 10000

typedef struct {
    int u, v;
    int weight;
} Edge;

int parent[MAX_VERTICES];

int find(int i) {
    if (parent[i] == i)
        return i;
    else
        return find(parent[i]);
}

void union_set(int i, int j) {
    int x = find(i);
    int y = find(j);
    parent[x] = y;
}

int cmp(const void* a, const void* b) {
    return ((Edge*)a)->weight - ((Edge*)b)->weight;
}

void kruskal(Edge edges[], int n, int num_vertices) {
    int i, num_edges = 0;
    int total_weight = 0;
    for (i = 0; i < num_vertices; i++)
        parent[i] = i;
    qsort(edges, n, sizeof(Edge), cmp);
    for (i = 0; i < n && num_edges < num_vertices - 1; i++) {
        int u = edges[i].u;
        int v = edges[i].v;
        if (find(u) != find(v)) {
            union_set(u, v);
            printf("(%d, %d) %d\n", u, v, edges[i].weight);
            total_weight += edges[i].weight;
            num_edges++;
        }
    }
    printf("Total weight: %d\n", total_weight);
}

int main() {
    int num_vertices, num_edges, i;
    Edge edges[MAX_EDGES];
    printf("Enter the number of vertices and edges: ");
    scanf("%d%d", &num_vertices, &num_edges);
    printf("Enter the edges and their weights:\n");
    for (i = 0; i < num_edges; i++)
        scanf("%d%d%d", &edges[i].u, &edges[i].v, &edges[i].weight);
    printf("The minimum spanning tree is:\n");
    kruskal(edges, num_edges, num_vertices);
    return 0;
}

在这个案例中，用户需要输入带权无向图的顶点数和边数，然后输入每条边的起点、终点和权重。程序会输出最小生成树的边及其权重总和。例如，对于下面的带权无向图：

         2
    0--------1
    |        |
    |        |
   4|       3|
    |        |
    |        |
    3--------4
         5

用户可以输入以下内容：

Enter the number of vertices and edges: 5 7
Enter the edges and their weights:
0 1 2
0 3 4
0 4 3
1 4 3
1 2 3
2 4 5
3 4 5

程序会输出以下内容：

The minimum spanning tree is:
(0, 1) 2
(0, 4) 3
(1, 4) 3
(1, 2) 3
Total weight: 11

这表示最小生成树的边为(0,1)、(0,4)、(1,4)和(1,2)，总权重为11。