洛谷P2024 食物链 [NOI2001] 并查集

正解:并查集

解题报告:

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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=+;
ll n,k,fa[N*],ans;

inline ll read()
{
    register char ch=getchar();register ll x=;register bool y=;
    while(ch!='-' && (ch>'' || ch<''))ch=getchar();
    if(ch=='-')ch=getchar(),y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),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,,n*)fa[i]=i;
    rp(i,,k)
    {
        ll op=read();
        if(op==)
        {
            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 ;
}
//fax:xÒ»Àà fax+n:x³Ôʲô fax+2n:x±»³Ô