题解

最后的结果与约束条件的顺序无关，可以先考虑相等条件，再考虑不等条件。由于题目中i和j的数据范围较大，需要用到离散化。

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

const int N = 200010;

int n, m;
int p[N];
unordered_map<int, int> S;

struct Query
{
    int x, y, e;
} query[N];

int get(int x)
{
    if (S.count(x) == 0) S[x] = ++n;
    return S[x];
}

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

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        n = 0;
        S.clear();
        scanf("%d", &m);
        for (int i = 0; i < m; i++)
        {
            int x, y, e;
            scanf("%d%d%d", &x, &y, &e);
            query[i] = {get(x), get(y), e};
        }
        
        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;
            }
        
        // 检查所有不等条件
        bool has_conflict = false;
        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)
                {
                    has_conflict = true;
                    break;
                }
            }
        
        if (has_conflict) puts("NO");
        else puts("YES");
    }
    
    return 0;
}

参考资料

1. AcWing算法提高课