写一段c++代码，假设无向图用邻接表表示，试写一个公开的成员函数，以深度优先遍历方式输出图中每个连通分量所含的顶点集合。 程序满足1、每个连通分量单独输出成一行； 2、函数返回值为图中连通分量个数。

很高兴为您提供代码解决方案，以下是C++代码：

#include <iostream>
#include <vector>
#include <stack>
using namespace std;

class Graph {
    int V; // 顶点个数
    vector<int> *adj; // 邻接表
public:
    Graph(int V) {
        this->V = V;
        adj = new vector<int>[V];
    }

    void addEdge(int v, int w) {
        adj[v].push_back(w);
        adj[w].push_back(v);
    }

    int DFSUtil(int v, bool visited[], vector<int> &comp) {
        visited[v] = true;
        comp.push_back(v);

        vector<int>::iterator i;
        for (i = adj[v].begin(); i != adj[v].end(); ++i)
            if (!visited[*i])
                DFSUtil(*i, visited, comp);

        return comp.size();
    }

    int connectedComponents() {
        bool *visited = new bool[V];
        for (int v = 0; v < V; v++)
            visited[v] = false;

        int count = 0; // 连通分量个数

        for (int v = 0; v < V; v++) {
            if (!visited[v]) {
                vector<int> comp;
                int size = DFSUtil(v, visited, comp);
                count++;
                for (int i = 0; i < comp.size(); i++)
                    cout << comp[i] << " ";
                cout << endl;
            }
        }

        return count;
    }
};

// 测试程序
int main() {
    Graph g(5);
    g.addEdge(1, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 4);

    cout << "Following are connected components" << endl;
    int count = g.connectedComponents();
    cout << "Number of connected components: " << count << endl;

    return 0;
}

输出结果：

Following are connected components
0 1 
2 3 4 
Number of connected components: 2

当然，您可以将无向图的邻接表表示替换为您的实现。

希望能对您有所帮助！