思路: 只有最多2n个变量,所以为每个变量分配编号,并用编号进行并查集

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<map>
#include<stdio.h>
using namespace std;
map<int, int> mp;//记录超级大数对应的编号

int f[2000200];

int fd(int x) {//查找
    return (f[x] == x) ? (x) : (f[x] = fd(f[x]));
}

void hb(int x, int y) {//合并(迫真命名
    int _x = fd(x), _y = fd(y);

    if (_x != _y) f[_x] = _y;

}

int cw = 0;

class nd {
public:
    int a, b;
    nd() {
        a = 0; b = 0;
    }
    nd(int _1, int _2) {
        a = _1; b = _2;
    }
};

nd com[2000200];

void nc(int x, int y) {
    com[cw++] = nd(x, y);
}

void m1() {
    cw = 0;
    mp.clear();

    int w = 1;
    int n;
    scanf("%d", &n);
    for (int i = 0; i <= 2*n + 10; i++) {
        f[i] = i;
    }
    
    int _1, _2, _3;
    for (int i = 0; i < n; i++) {
        scanf("%d%d%d", &_1, &_2, &_3);
        if (_3) {//加入:为新数据分配一个编号,并用编号进行并查集
            map<int, int>::iterator l = mp.find(_1);
            map<int, int>::iterator r = mp.find(_2);
            if (l==mp.end()) {
                mp[_1] = w++;
                l = mp.find(_1);
            }
            if (r == mp.end()) {
                mp[_2] = w++;
                r = mp.find(_2);
            }
            //printf("1\n");
            hb(l->second, r->second);
        }
        else {
            if (_1 == _2) {//自己和自己不同就干掉
                printf("NO\n"); 
            
            return;
            }
            nc(_1, _2);
        }
    }

    for (int i = 0; i < cw; i++) {
        
        map<int, int>::iterator l2 = mp.find(com[i].a);
        map<int, int>::iterator r2 = mp.find(com[i].b);
        if (l2 == mp.end() || r2 == mp.end()) continue;
        //printf("2\n");
        if (fd(l2->second) == fd(r2->second)) {//干掉!
            printf("NO\n"); return;
        }
    }

    printf("YES\n");
    return;


}

int main() {
    int t;
    //freopen("1.in", "r", stdin);//好大的测试数据啊
    scanf("%d", &t);
    for (int i = 0; i < t; i++) {
        m1();
    }


    return 0;
}