正解:并查集
解题报告:
这题，大概难点在分类讨论？（划
反正整个儿就并查集板子题了
说难点在分类讨论主要是因为,一般这种都是分成两组嘛,就很简单fa[x]=y fa[x+n]=y+n差不多这意思嘛
但是这个是分成三组,,,所以打起来就要注意一下理清条理,最好注释下什么东西是干什么的,这样打起来其实还是挺susi的(就像之前做这个一样(逃
没啦!代码放最后辣!over!
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rp(i,x,y) for(register ll i=x;i<=y;++i)
#define lowbit(x) x&(-x)
const ll N=50000+100;
ll n,k,fa[N*3],ans;
inline ll read()
{
register char ch=getchar();register ll x=0;register bool y=1;
while(ch!='-' && (ch>'9' || ch<'0'))ch=getchar();
if(ch=='-')ch=getchar(),y=0;
while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=getchar();
return y?x:-x;
}
inline ll fd(ll x){return fa[x]==x?fa[x]:fa[x]=fd(fa[x]);}
int main()
{
n=read();k=read();rp(i,1,n*3)fa[i]=i;
rp(i,1,k)
{
ll op=read();
if(op==1)
{
ll x=read(),y=read();
if(x>n || y>n){++ans;continue;}
ll fa1=fd(x),fa2=fd(y),fa3=fd(x+n),fa4=fd(y+n),fa5=fd(x+n+n),fa6=fd(y+n+n);
if(fa3==fa2 || fa5==fa2){++ans;continue;}
fa[fa1]=fa2;fa[fa3]=fa4;fa[fa5]=fa6;
continue;
}
ll x=read(),y=read();
if(x>n || y>n || x==y){++ans;continue;}
ll fa1=fd(x),fa2=fd(y),fa3=fd(x+n),fa4=fd(y+n),fa5=fd(x+n+n),fa6=fd(y+n+n);
if(fa1==fa2 || fa5==fa2){++ans;continue;}
fa[fa3]=fa2;fa[fa6]=fa1;fa[fa5]=fa4;
}
printf("%lld\n",ans);
return 0;
}
//fax:xÒ»Àà fax+n:x³Ôʲô fax+2n:x±»³Ô