图的深度和广度优先遍历

题目描述
以邻接矩阵给出一张以整数编号为顶点的图，其中0表示不相连，1表示相连。按深度和广度优先进行遍历，输出全部结果。要求，遍历时优先较小的顶点。如，若顶点0与顶点2，顶点3，顶点4相连，则优先遍历顶点2.

输入
顶点个数
邻接矩阵

输出
DFS
深度遍历输出
WFS
广度遍历输出,

样例输入
3
0 1 1
1 0 1
1 1 0

样例输出
DFS
0 1 2
1 0 2
2 0 1
WFS
0 1 2
1 0 2
2 0 1
 

#include <iostream>
#include <queue>

using namespace std;
int* visit;
int num;
int** edge;

void DFS(int n) {
    visit[n] = 1;
    cout << n << " ";
    for (int i = 0; i <num; i++) {//判断领顶点
        if (edge[n][i] == 1 && visit[i] == 0)
            DFS(i);
    }
}

void BFS(int n) {
    int temp = n;
    queue<int>q;
    q.push(temp);

    while (!q.empty()) {
        int t = q.front();
        q.pop();
        if (visit[t] == 0) {//没访问过就输出
            cout << t << " ";
            visit[t] = 1;
        }

        for (int i = 0; i <num; i++) {//把他的领顶点放到队中
            if (edge[t][i] == 1 && visit[i] == 0) {
                q.push(i);
            }
        }
    }
}

void Reset() {
    for (int i = 0; i < num; i++)
        visit[i] = 0;

    cout << endl;
}

int main() {
    cin >> num;

//判断是否走过
    visit = new int[num];
    for (int i = 0; i < num; i++)
        visit[i] = 0;

//邻接矩阵
    edge = new int* [num];
    for (int i = 0; i < num; i++)
        edge[i] = new int[num];
    for (int i = 0; i < num; i++)
        for (int j = 0; j < num; j++)
            cin >> edge[i][j];


    cout << "DFS" << endl;;
    for (int i = 0; i < num; ++i) {
        DFS(i);
        Reset();
    }
    cout << "WFS" << endl;;
    for (int i = 0; i < num; ++i) {
        BFS(i);
        Reset();
    }

}