原题链接
题解
关系具有矢量特性,因此可以带权并查集维护
code
#include<bits/stdc++.h>
#define ll long long
using namespace std;int fa[50006];
int val[50006];
int finds(int now)
{if(now==fa[now]) return now;int tem=fa[now];fa[now]=finds(fa[now]);val[now]=(val[now]+val[tem])%3;return fa[now];
}void solve()
{int n,k;cin>>n>>k;for(int i=1;i<=n;i++){fa[i]=i;val[i]=0;}int ans=0;for(int i=1;i<=k;i++){int op,x,y;cin>>op>>x>>y;if(x>n||y>n||op==2&&x==y){ans++;continue;}if(op==1){int fx=finds(x),fy=finds(y);if(fx==fy){if(val[x]!=val[y]) ans++;continue;}int tem=val[x];fa[fx]=fy;val[fx]=(3-tem+val[y])%3;}else{int fx=finds(x),fy=finds(y);if(fx==fy){if((val[x]-val[y]+3)%3!=1) ans++;continue;}int tem=val[x];fa[fx]=fy;val[fx]=(3-tem+val[y]+1)%3;}}cout<<ans;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t=1;//cin>>t;while(t--) solve();return 0;
}