赛时写的神奇代码，大致思路是分成两组，然后只要有冲突就丢尽另外一组，如果还有冲突就判断为 NO。但发现会有两组都可以放的情况，我直接把它丢到队列里等一轮完后再次去判断，如果连续丢到队列的次数过多，就将它放入第一组。

这种神奇的贪心思路竟然过了，而且赛后也没有被 hack，请求大佬能够把它叉了 qwq。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#define init(x) memset (x,0,sizeof (x))
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
using namespace std;
const int MAX = 2e5 + 5;
const int MOD = 1e9 + 7;
inline int read ();
int t,n,cnt,a[MAX],b[MAX];
bool ok; 
int main ()
{
	//freopen (".in","r",stdin);
	//freopen (".out","w",stdout);
	t = read ();
	while (t--)
	{
		n = read ();
		ok = 1;
		queue <pair <int,int> > q;
		for (int i = 1;i <= n;++i)
		{
			int x = read (),y = read ();
			a[i] = b[i] = 0;
			q.push ({x,y});
		}
		while (!q.empty ())
		{
			int x = q.front ().first,y = q.front ().second;
			q.pop ();
			if (x == y) ok = 0;
			else if ((a[x] || a[y]) && (b[x] || b[y])) ok = 0;
			else if (!a[x] && !a[y] && (b[x] || b[y])) a[x] = a[y] = 1;
			else if (!b[x] && !b[y] && (a[x] || a[y])) b[x] = b[y] = 1;
			else if (cnt >= 10000) a[x] = a[y] = 1,cnt = 0;//就是这里比较神奇 赛后改为1000就过不了了
			else ++cnt,q.push ({x,y});
		}
		if (ok) printf ("YES\n");
		else printf ("NO\n");
	}
	return 0;
}
inline int read ()
{
    int s = 0;int f = 1;
    char ch = getchar ();
    while ((ch < '0' || ch > '9') && ch != EOF)
	{
        if (ch == '-') f = -1;
        ch = getchar ();
    }
    while (ch >= '0' && ch <= '9')
	{
        s = s * 10 + ch - '0';
        ch = getchar ();
    }
    return s * f;
}