帮我用c语言写一个用克鲁斯卡尔算法写的最小生成树

好的，下面是用C语言实现的Kruskal算法求最小生成树的代码：

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

#define MAX_VERTEX_NUM 20 // 最大顶点数
#define MAX_EDGE_NUM 50 // 最大边数

typedef struct {
    int u, v; // 边的两个顶点
    int weight; // 边权值
} Edge;

typedef struct {
    int vertex[MAX_VERTEX_NUM]; // 存储顶点
    Edge edge[MAX_EDGE_NUM]; // 存储边
    int edgeNum; // 边的数量
    int vertexNum; // 顶点的数量
} Graph;

void initGraph(Graph* graph, int vertexNum) {
    graph->edgeNum = 0;
    graph->vertexNum = vertexNum;
    for (int i = 0; i < vertexNum; i++) {
        graph->vertex[i] = i;
    }
}

int find(int parent[], int i) {
    while (parent[i] != i) {
        i = parent[i];
    }
    return i;
}

void unionSet(int parent[], int x, int y) {
    int xRoot = find(parent, x);
    int yRoot = find(parent, y);
    parent[xRoot] = yRoot;
}

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

void kruskal(Graph* graph) {
    int parent[MAX_VERTEX_NUM];
    for (int i = 0; i < graph->vertexNum; i++) {
        parent[i] = i;
    }
    qsort(graph->edge, graph->edgeNum, sizeof(Edge), cmp);
    for (int i = 0; i < graph->edgeNum; i++) {
        int u = graph->edge[i].u;
        int v = graph->edge[i].v;
        int weight = graph->edge[i].weight;
        int uRoot = find(parent, u);
        int vRoot = find(parent, v);
        if (uRoot != vRoot) {
            printf("Edge %d:(%d, %d) weight: %d\n", i, u, v, weight);
            unionSet(parent, u, v);
        }
    }
}

int main() {
    Graph graph;
    int vertexNum = 6;
    initGraph(&graph, vertexNum);
    graph.edgeNum = 9;
    graph.edge[0].u = 0;
    graph.edge[0].v = 1;
    graph.edge[0].weight = 1;
    graph.edge[1].u = 0;
    graph.edge[1].v = 2;
    graph.edge[1].weight = 6;
    graph.edge[2].u = 0;
    graph.edge[2].v = 3;
    graph.edge[2].weight = 5;
    graph.edge[3].u = 1;
    graph.edge[3].v = 2;
    graph.edge[3].weight = 3;
    graph.edge[4].u = 1;
    graph.edge[4].v = 4;
    graph.edge[4].weight = 2;
    graph.edge[5].u = 2;
    graph.edge[5].v = 3;
    graph.edge[5].weight = 4;
    graph.edge[6].u = 2;
    graph.edge[6].v = 4;
    graph.edge[6].weight = 7;
    graph.edge[7].u = 3;
    graph.edge[7].v = 5;
    graph.edge[7].weight = 9;
    graph.edge[8].u = 4;
    graph.edge[8].v = 5;
    graph.edge[8].weight = 8;
    kruskal(&graph);
    return 0;
}

这份代码使用了并查集来判断两个顶点是否在同一个连通分量中，从而避免了生成环。时间复杂度为O(ElogE)，其中E为边的数量。