使用c语言写一个代码假设无向连通图采用邻接表存储。设计一个算法输出图G的一颗深度优先生成树

好的，对于您的问题，我可以提供以下 C 语言代码来实现：

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

// 定义邻接表结点
typedef struct node {
    int vertex;
    struct node* next;
} Node;

// 定义邻接表
typedef struct adj_list {
    Node* head;
} AdjList;

// 定义图结构
typedef struct graph {
    int vertices;
    AdjList* array;
} Graph;

// 初始化邻接表
Node* create_node(int vertex) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->vertex = vertex;
    newNode->next = NULL;
    return newNode;
}

// 创建图
Graph* create_graph(int vertices) {
    Graph* graph = (Graph*)malloc(sizeof(Graph));
    graph->vertices = vertices;
    graph->array = (AdjList*)malloc(vertices * sizeof(AdjList));
    int i;
    for (i = 0; i < vertices; ++i) {
        graph->array[i].head = NULL;
    }
    return graph;
}

// 添加边
void add_edge(Graph* graph, int src, int dest) {
    // 添加边（src -> dest）
    Node* newNode = create_node(dest);
    newNode->next = graph->array[src].head;
    graph->array[src].head = newNode;

    // 添加边（dest -> src），由于无向图的存在，需要添加两条边
    newNode = create_node(src);
    newNode->next = graph->array[dest].head;
    graph->array[dest].head = newNode;
}

// 深度优先搜索
void dfs(int vertex, int* visited, Graph* graph) {
    visited[vertex] = 1;
    printf("%d ", vertex);

    // 遍历顶点的每个邻接顶点
    Node* adj_vertex = graph->array[vertex].head;
    while (adj_vertex != NULL) {
        if (!visited[adj_vertex->vertex]) {
            dfs(adj_vertex->vertex, visited, graph);
        }
        adj_vertex = adj_vertex->next;
    }
}

// 输出图 G 的一颗深度优先生成树
void dfs_spanning_tree(Graph* graph) {
    int* visited = (int*)malloc(graph->vertices * sizeof(int));
    int i;
    for (i = 0; i < graph->vertices; ++i) {
        visited[i] = 0;
    }

    // 从第一个顶点开始遍历
    dfs(0, visited, graph);
}

// 主函数
int main() {
    int vertices = 5;
    Graph* graph = create_graph(vertices);
    add_edge(graph, 0, 1);
    add_edge(graph, 0, 2);
    add_edge(graph, 1, 2);
    add_edge(graph, 1, 3);
    add_edge(graph, 2, 3);
    add_edge(graph, 3, 4);

    dfs_spanning_tree(graph);

    return 0;
}

希望这可以帮助您解决问题！