并查集详解
https://blog.csdn.net/sjystone/article/details/115406797

程序自动分析 https://www.acwing.com/problem/content/239/

离散化 + 并查集

离散化：
1）有序：排序 去重 二分
2）无序：哈希表 / map

将相等的条件放到一个集合中
遍历不相等条件，若在一个集合中则不满足

#include <iostream>
#include <unordered_map>
using namespace std;

const int N = 1000010;
int p[N], n, cnt, T;
unordered_map<int, int> mp;

struct Query{
    int a, b, e;
} query[N];

int find(int x)
{
    if ( x != p[x] ) p[x] = find(p[x]);
    return p[x];
}

int get(int x)
{
    if ( !mp.count(x) ) mp[x] = ++ cnt; //离散(无序)
    return mp[x];
}

int main()
{
    cin >> T;
    while ( T -- )
    {
        cin >> n;
        cnt = 0;
        bool judge = false;
        mp.clear();
        for ( int i = 1; i <= n; i ++ )
        {
            int a, b, e;
            cin >> a >> b >> e;
            query[i] = {get(a), get(b), e};
        }
        for ( int i = 1; i <= cnt; i ++ ) p[i] = i; //到cnt，易错
        for ( int i = 1; i <= n; i ++ )
            if ( query[i].e == 1 )
            {
                int a = query[i].a, b = query[i].b;
                int pa = find(a), pb = find(b);
                p[pa] = pb;
            }
        for ( int i = 1; i <= n; i ++ )
            if ( query[i].e == 0 )
            {
                int a = query[i].a, b = query[i].b;
                if ( find(a) == find(b) )
                {
                    judge = true;
                    break;
                }
            }
        if ( judge ) cout << "NO" << endl;
        else cout << "YES" << endl;
    }
    return 0;
}