1272小希的迷宫

思路是用并查集判断连通，然后判断边数+1==点数，就是判断是否为一颗生成树

#include <iostream>
#include <set>
#include <algorithm>
using namespace std;

#ifndef ONLINE_JUDGE
#include <fstream>
ifstream fin("test.txt");
#define cin fin
#endif
const int MAXN = 100010;
int p[MAXN],vis[MAXN],maxv,edge,ok;
set<int> s;
int find(int x)
{
    return x == p[x] ? x : p[x] = find(p[x]);
}
void ini()
{
    ok = 1;
    s.clear();
    maxv = edge = 0;
    memset(vis,0,sizeof(vis));
    for(int i = 0; i < 100000; ++i)
    p[i] = i;
}
int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    ini();
    while(cin >> n >> m)
    {
        if(n == -1 && -1 == m)
        break;
        maxv = max(m,max(maxv,n));
        if(!n && !m)
        {
            if(edge == 0)     //如果只有2个0的情况，直接输出yes
            {
                cout << "Yes" << endl;
                continue;
            }
            if(s.size()!=edge+1)
            {
                cout << "No" << endl;
                ini();
                continue;
            }
            int cnt = 0;
            for(int i = 1; i <= maxv; ++i)
            if(vis[i] && p[i] == i)
            cnt++;
            if(cnt == 1 && ok)
            cout << "Yes" << endl;
            else
            cout << "No" << endl;
            ini();
            continue;
        }
        s.insert(n);
        s.insert(m);
        edge++;
        vis[n] = vis[m] = 1;
        n = find(n);
        m = find(m);
        if(n != m)
        p[m] = n;
        else 
        ok = 0;    //如果2个点在同一个集合，直接判否，因为构成了回路
    }
    return 0;
}