帮我用c语言写一个代码，要求输入无向连通图顶点数与边数，并每行输入各边的顶点及其权值数据（顶点序号从0开始），采用Kruskal算法求解该无向连通图的最小生成树依久得到的边，并计算输出其权值。

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

typedef struct Edge {
    int src, dest, weight;
} Edge;

typedef struct Graph {
    int vertices;
    Edge** edges;
    bool* visited;
} Graph;

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

void kruskal_mst(Graph* graph) {
    int i, j, edge_count = 0;
    qsort(graph- 1) / 2, sizeof(Edge), compare);

    for (i = 0; i < graph->vertices && edge_count < graph->vertices - 1; i++) {
        if (!graph->visited[i] && !graph->visited[(Edge*)graph->edges[edge_count].dest]) {
            printf("Edge %d-%d with weight %d\n", graph->edges[edge_count].src, graph->edges[edge_count].dest, graph->edges[edge_count].weight);
            graph->visited[i] = true;
            graph->visited[(Edge*)graph->edges[edge_count].dest] = true;
            edge_count++;
        }
    }
}

int main() {
    int v, e, src, dest, weight;
    printf("Enter number of vertices and edges: ");
    scanf("%d %d", &v, &e);

    Graph* g = (Graph*) malloc(sizeof(Graph));
    g->vertices = v;
    g->edges = (Edge**)malloc(v * (v - 1) * sizeof(Edge*));
    g->visited = (bool*)malloc(v * sizeof(bool));

    // Initialize visited array to false
    for (int i = 0; i < v; i++)
        g->visited[i] = false;

    for (int i = 0; i < e; i++) {
        printf("Enter edge (%d, %d) with weight: ", i, i + 1); // Assuming edges are undirected, input symmetrically
        scanf("%d %d %d", &src, &dest, &weight);
        Edge* edge = (Edge*)malloc(sizeof(Edge));
        edge->src = src;
        edge->dest = dest;
        edge->weight = weight;
        g->edges[i] = edge;
    }

    kruskal_mst(g);

    free(g->edges);
    free(g->visited);
    free(g);
    return 0;
}