小希的迷宫

#pragma comment(linker, "/STACK:102400000,102400000")  // maybe爆栈，有两种解决办法，第一种加

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

#define N 100005

int f[N], use[N];

void init()
{
    memset(use, 0, sizeof(use));

    for(int i = 0; i < N; i++)
        f[i] = i;
}
int found(int x)
{
    //int k = f[x];
    if(f[x] != x)
    {
        f[x] = found(f[x]);
        //r[x] = r[x]+r[k];
    }
    return f[x];
}

int main()
{
    int a, b, sum = 0;
    init();
    int flag = 0;

    while(1)
    {
        cin >> a >> b;

        if(a == -1 || b == -1)
            break;


        if(a+b == 0)
        {
            sum = 0;
            for(int i = 0; i < N; i++)
                if(use[i] && f[i] == i)
                    sum++;

            if(flag == 0 && sum < 2)  // 只准有一棵树，一个根节点
                cout << "Yes" << endl;
            else
                cout << "No" << endl;

            init();
            flag = 0;
        }
        else
        {
            use[a] = use[b] = 1;
            int nx = found(a), ny = found(b);
            if(nx != ny)
                f[nx] = ny;
            else flag = 1;  // 如果两个房间已经有联系了，这条路就没有价值，就不满足小希的思想
        }
    }
    return 0;
}

#include<iostream>
#include<cstring>

using namespace std;

#define N 100005

bool vis[N];

int main()
{
    int a, b, k = 0, m = 0;
    memset(vis, false, sizeof(vis));

    while(cin >> a >> b, a != -1 || b != -1)
    {
        if(a+b == 0)
        {
            for(int i = 0; i < N; i++)
            {
                if(vis[i])
                    m++;  // 计算房间出现的数量
            }
            if(m - k == 1 || m == 0)   // 如果满足小希的思想话，房间数减去联通的路数一定等于1.因为每一条路都连着两个房间，而且每条路都有价值。如果输入两个0的话，没有房间，还输出Yes
                cout << "Yes" << endl;
            else
                cout << "No" << endl;
            k = 0;
            m = 0;
            memset(vis, false, sizeof(vis));
        }
        else
        {
            vis[a] = vis[b] = true;   // 出现的房间数置为true
            k++;  // k记录路数
        }
    }
    return 0;
}