Description

给定多个 $ x_i  x_j $ 是否相等的条件

判断能否实现给每个 $  x_ i  $赋上合适的值满足条件

Solution

考虑用并查集实现

若两个数相等，则表示它们的祖先相同

给出的条件要先排序，把所有相同的条件放在前面先处理

数的范围很大，并查集数组开不下，需要离散化一下

Code

#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int T, n, f[N], hash[N], tot = 0; 

struct Node {
    int a, b, c;
}u[N];

int getf(int u) {
    return u == f[u] ? u : f[u] = getf(f[u]);
}


template <typename T>
void read(T &t) {
    t = 0; T m = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') m = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { t = (t << 3) + (t << 1) + (ch & 15); ch = getchar(); }
    t *= m;
}

bool cmp(const Node &a, const Node &b) {
    return a.c > b.c;
}

int main() {
    read(T);
    while(T--) {
        tot = 0;
        memset(f, 0, sizeof(f));
        read(n);
        bool fl = true;
        for(register int i = 1; i <= n; i++) {
            read(u[i].a), read(u[i].b), read(u[i].c);
            hash[++tot] = u[i].a;
            hash[++tot] = u[i].b;
        }
        sort(hash + 1, hash + tot + 1);
        int siz = unique(hash + 1, hash + tot + 1) - hash - 1;
        for(register int i = 1; i <= n; i++) {
            u[i].a = lower_bound(hash + 1, hash + siz + 1, u[i].a) - hash;
            u[i].b = lower_bound(hash + 1, hash + siz + 1, u[i].b) - hash;
        }
        for(register int i = 1; i <= siz * 2; i++) f[i] = i;
        sort(u + 1, u + n + 1, cmp);
        for(register int i = 1; i <= n; i++) {
            int A = u[i].a, B = u[i].b, C = u[i].c;
            int F1 = getf(A), F2 = getf(B);
            if(C == 0) {
                if(F1 == F2) { fl = false; break; }
            } else {
                if(F1 != F2) f[F1] = f[F2];
            }
        }
        printf("%s\n", fl ? "YES" : "NO");
    }
    return 0;
}