离散化+并查集

关系的化就是并查集处理，但是数据规模很大，下标到了1e9，所以开不下那么大的并查集数组，那么怎么办，离散化去搞。
我们不需要知道每个数具体多大，只需要知道相对大小，找到这个数就行了。
别用set，常数很大，就用快排+去重。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
int x[N], y[N], op[N];
int a[N << 1], fa[N << 1];

int find(int x) {
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}

int main() {
    int T = read();
    while (T--) {
        for (int i = 1; i < N << 1; ++i)    fa[i] = i;
        int n = read();
        for (int i = 1; i <= n; ++i) {
            x[i] = read(), y[i] = read(), op[i] = read();
            a[i * 2 - 1] = x[i], a[i * 2] = y[i];
        }
        int m = n << 1;
        sort(a + 1, a + 1 + m);
        m = unique(a + 1, a + 1 + m) - a - 1;
        for (int i = 1; i <= n; ++i) {
            int dx = lower_bound(a + 1, a + 1 + m, x[i]) - a;
            int dy = lower_bound(a + 1, a + 1 + m, y[i]) - a;
            if (op[i])    fa[find(dx)] = find(dy);
        }
        int flag = 0;
        for (int i = 1; i <= n; ++i) {
            int dx = lower_bound(a + 1, a + 1 + m, x[i]) - a;
            int dy = lower_bound(a + 1, a + 1 + m, y[i]) - a;
            if (!op[i] and fa[find(dx)] == fa[find(dy)]) {
                flag = 1;    break;
            }
        }
        if (flag)    puts("NO");
        else puts("YES");
    }
    return 0;
}