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

食物链 https://www.acwing.com/problem/content/242/

每次合并集合a，b的时候
用d[pa]维护a和b的关系

若a，b在一个集合
    若(d[a] - d[b]) % 3 = 1 -> a吃b
    若(d[a] - d[b]) % 3 = 0 -> ab同类
若a，b不在同一集合
    若a吃b，则令d[pa] = d[b] - d[a] + 1 -> 以满足(d[a] + d[pa] - d[b]) % 3 = 1
    若ab同类，则令d[pa] = d[b] - d[a] -> 以满足(d[a] + d[pa] - d[b]) % 3 = 0;

#include <iostream>
using namespace std;

const int N = 1e5 + 10;
int p[N], d[N], n, m, res;

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

int main()
{
    cin >> n >> m;
    for ( int i = 1; i <= n; i ++ ) p[i] = i;
    while ( m -- )
    {
        int op, a, b;
        cin >> op >> a >> b;
        if ( a > n || b > n ) res ++;
        else 
        {
            int pa = find(a), pb = find(b);
            if ( op == 1 )
            {
                if ( pa == pb && (d[a] - d[b]) % 3 ) res ++;
                else if ( pa != pb ) 
                {
                    p[pa] = pb;
                    d[pa] = d[b] - d[a];
                }
            }
            else
            {
                if ( pa == pb && (d[a] - d[b] - 1) % 3 ) res ++;
                else if ( pa != pb )
                {
                    p[pa] = pb;
                    d[pa] = d[b] - d[a] + 1;
                }
            }
        }
    }
    cout << res;
    return 0;
}