正解:并查集
解题报告:
传送门
其实没有很难(虽然我是交了三发才过的QAQ
但是一来好久没打并查集了恢复一下智力
二来看着智推里唯一一个蓝就很不爽(,,,虽然做了这题之后又补上了个蓝题QAQ
所以就做了下QwQ
先放下代码趴QAQ明天写下解法,难点在于分类讨论
#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