P2024 [NOI2001]食物链[扩展域并查集]

大水题一道啊，几分钟切掉。

还是扩展域，每个点拆3个点，之间连边表示有关系（即捕食关系）。然后随便判定一下就好了，不难，毕竟NOI上古题目。

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cmath>

 #include<algorithm>

 #include<queue>

 #define dbg(x) cerr<<#x<<" = "<<x<<endl

 #define ddbg(x,y) cerr<<#x<<" = "<<x<<"   "<<#y<<" = "<<y<<endl

 using namespace std;

 typedef long long ll;

 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,:;}

 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,:;}

 template<typename T>inline T _min(T A,T B){return A<B?A:B;}

 template<typename T>inline T _max(T A,T B){return A>B?A:B;}

 template<typename T>inline T read(T&x){

     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;

     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;

 }

 const int N=+;

 int fa[N<<],n,m,x,y,c,cnt;

 inline int Get(int x){return fa[x]^x?fa[x]=Get(fa[x]):x;}

 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);

     read(n),read(m);

     for(register int i=;i<=n*;++i)fa[i]=i;

     for(register int i=;i<=m;++i){

         read(c),read(x),read(y);

         if(x>n||y>n){++cnt;continue;}

         if(c==){

             if(Get(x)==Get(y+n)||Get(x)==Get(y+n+n)){++cnt;continue;}

             fa[Get(x)]=Get(y),fa[Get(x+n)]=Get(y+n),fa[Get(x+n+n)]=Get(y+n+n);

         }

         else{

             if(Get(x)==Get(y)||Get(x)==Get(y+n+n)){++cnt;continue;}

             fa[Get(x)]=Get(y+n),fa[Get(x+n)]=Get(y+n+n),fa[Get(x+n+n)]=Get(y);

         }

     }

     printf("%d\n",cnt);

     return ;

 }

P2024 [NOI2001]食物链[扩展域并查集]的相关教程结束。