DFS---迷宫问题

#include<iostream>
#include<string>
#include<cstring>
using namespace std;//dfs
int flag;
int p[3];
int a[200][200];
void Dfs(int n,int m)//起点,终点
{
    if (n == m)
    {
        flag = 1;
        return;
    }
    for (int i = 0; i < 3; i++)
    {
        if (a[n][i] && !p[a[n][i]])
        {
            p[a[n][i]] = 1;
            Dfs(a[n][i], m);
            p[a[n][i]] = 0;//???
        }
    }
}
int main()
{
    int n;
    int m;
    while (cin >> n && n != 0)
    {
        flag = 0;
        for (int i = 1; i <=n;i++)
        {
            for (int j = 0; j < 3; j++)
            {
                cin >> a[i][j];
            }
        }
        cin >> m;//代表终点
        p[1] = 1;//起点访问过
        Dfs(1,m);
        if (flag == 1)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;

    }
    return 0;
}
原文地址:https://www.cnblogs.com/h694879357/p/11771923.html