06-图1 列出连通集 (25分)

给定一个有NN个顶点和EE条边的无向图，请用DFS和BFS分别列出其所有的连通集。假设顶点从0到N-1N−1编号。进行搜索时，假设我们总是从编号最小的顶点出发，按编号递增的顺序访问邻接点。

输入格式:

输入第1行给出2个整数NN(0

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

class Graph {
private:
int Nv;             //顶点数
int Ne;             //边数
int** G;            //邻接矩阵
int* isVisited; //访问数组
public:
//构造函数
Graph(int nv,int ne) {
this->Nv = nv;
this->Ne = ne;
this->isVisited = new int[nv];
this->G = new int*[nv];
for ( int i = 0 ; i < nv ; i++) {
this->G[i] = new int[nv];
this->isVisited[i] = 0;
}
for ( int i = 0 ; i  < nv ; i++){
for ( int j = 0 ; j < nv ; j++){
this->G[i][j] = 0;
this->G[i][i] = 1;
}
}

for ( int i = 0 ; i < this->Ne ; i++){
int a,b;
cin>>a>>b;
this->G[a][b] = 1;
this->G[b][a] = 1;
}
}

//深度优先遍历
void DFS(int start) {
cout<<" "<<start;
this->isVisited[start] = 1;
for ( int i = 0 ; i < this->Nv ; i++) {
if ( this->G[start][i] && !this->isVisited[i]) {
DFS(i);
}
}
}

void BFS(int start) {
queue<int> que;
this->isVisited[start] = 1;
que.push(start);
while(!que.empty()) {
int neiborgh = que.front();
cout<<" "<<neiborgh;
que.pop();
for ( int i = 0 ; i < this->Nv ; i++) {
if ( this->G[neiborgh][i] && !this->isVisited[i]) {
que.push(i);
this->isVisited[i] = 1;
}
}
}
}

//列出DFS连通集
void ListCompentDFS() {
for ( int i = 0 ; i < this->Nv ; i++){
if ( this->isVisited[i] == 0){
cout<<"{";
DFS(i);
cout<<" }"<<endl;
}
}
}

//列出BFS连通集
void ListCompentBFS() {
queue<int> que;
for ( int i = 0 ; i < this->Nv ; i++){
if (this->isVisited[i] == 0){
cout<<"{";
BFS(i);
cout<<" }"<<endl;
}
}
}

void Memset_Visited() {
for ( int i = 0 ; i < this->Nv ; i++){
this->isVisited[i] = 0;
}
}
};

int main() {
int nv,ne;
cin>>nv>>ne;
Graph graph(nv,ne);

graph.ListCompentDFS();
graph.Memset_Visited();
graph.ListCompentBFS();

return 0;
}