先把相等条件的放在一个集合中，在考虑不等条件，出现不等条件的祖宗节点是同一个就会有矛盾

import java.util.*;
import java.io.*;

public class Main {
    static final int N = 2000010;//取数据的2倍
    static int[] p = new int[N];
    static Query[] query = new Query[N];
    static HashMap<Integer, Integer> hash = new HashMap<>();
    static int n, m;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        int T = Integer.parseInt(br.readLine());        
        while (T -- > 0) {
            n = 0;//idx标记
            hash.clear();//清空hash进行下一轮测试
            m = Integer.parseInt(br.readLine());
            for (int i = 0; i < m; i++) {
                String[] str = br.readLine().split(" ");
                int x = Integer.parseInt(str[0]), 
                    y = Integer.parseInt(str[1]), 
                    e = Integer.parseInt(str[2]);
                //先离散化在加入query
                query[i] = new Query(get(x), get(y), e);
            }
            //初始化并查集
            for (int i = 1; i <= n; i++) p[i] = i;
            //合并所有相等约束条件
            for (int i = 0; i < m; i++) 
                if (query[i].e == 1) {
                    int pa = find(query[i].x), pb = find(query[i].y);
                    p[pa] = pb;
                }

            //检查所有的不等条件
            boolean flag = true;
            for (int i = 0; i < m; i++) 
                if (query[i].e == 0) {
                    int pa = find(query[i].x), pb = find(query[i].y);
                    if (pa == pb) {
                        flag = false;
                        break;
                    }
                }
            if (flag) out.println("YES");
            else out.println("NO");
        }
        out.flush();
    }

    static int find(int u) {
        if (u != p[u]) p[u] = find(p[u]);
        return p[u];
    }
    /*
    不难发现该离散化的value就是插入顺序
    本题并不要求离散化的数据有序性，对于有序性的
    离散化可以使用排序去重加二分操作，
    或者对于hash表的value进行操作
    */
    static int get(int x) {
        if (!hash.containsKey(x)) hash.put(x, ++n);
        return hash.get(x);
    }
}

class Query {
    int x, y, e;

    public Query(int x, int y, int e) {
        this.x = x;
        this.y = y;
        this.e = e;
    }
}