食物链（并查集）

一开始默认为0，如果有捕食关系调整d[x]

#include<bits/stdc++.h>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
vector<int> vx;
inline void divide() {sort(vx.begin(),vx.end());vx.erase(unique(vx.begin(),vx.end()),vx.end());}
inline int mp(int x) {return upper_bound(vx.begin(),vx.end(),x)-vx.begin();}
inline int log_2(int x) {return 31-__builtin_clz(x);}
inline int popcount(int x) {return __builtin_popcount(x);}
inline int lowbit(int x) {return x&-x;}
const int N = 50100;
//d[i]表示i到根节点的异或值是1还是0
int f[N],d[N];
int find(int u)
{
    if(f[u] == u) return u;
    int t = find(f[u]);
    d[u] += d[f[u]], d[u]%=3;
    return f[u] = t;
}
void merge(int u,int v,int st)
{
    int f1 = find(u), f2 = find(v);
    //要实现的效果为(d[f1]+d[u])%3 = (d[v]+st)%3
    d[f1] = (d[v]+st-d[u]+3)%3, f[f1] = f2;
}
void solve()
{
    int n,k;
    cin>>n>>k;
    int cnt = 0;
    for(int i=1;i<=n;++i) f[i] = i;
    for(int i=0;i<k;++i)
    {
        int st,u,v;
        cin>>st>>u>>v;
        if(u>n||v>n||u==v&&st==2) cnt++;
        else if(find(u) == find(v))
        {
            if((d[u]-d[v]+3)%3 != st-1) 
            {
                //cout<<u<<' '<<v<<' '<<d[u]<<' '<<d[v]<<'\n';
                cnt++;
            }
        }
        else
        {
            merge(u,v,st-1);
        }
    }
    cout<<cnt<<'\n';
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T = 1;
	//cin>>T;
	while(T--)
	{
		solve();
	}
}