【LOJ#3197】【eJOI2019】—T 形覆盖（并查集）

传送门

考虑2个关键点之间的关系

如果挨在一起，那么周围选的6个是确定的

如果角挨在一起，那么周围选的6个也是确定的

如果隔了一个格子，那么就是在周围7个里面去掉最小的一个

维护相关联的一堆关键点
看周围能填的格子数是否足够

#include<bits/stdc++.h>
using namespace std;
const int RLEN=(1<<20)|5;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define pb push_back
#define re register
#define fi first
#define se second
#define pii pair<int,int>
#define cs const
#define bg begin
#define ll long long
inline ll readl(){
	char ch=gc();
	ll res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
inline int add(int a,int b){a+=b-mod;return a+(a>>31&mod);}
inline void Add(int &a,int b){a+=b-mod;a+=a>>31&mod;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline void Dec(int &a,int b){a-=b;a+=a>>31&mod;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
cs int N=2000005;
int m,n,q;
int pos(int x,int y){
	return x*n+y;
}
int vis[N];
int a[N],sum,fa[N],tot,ans;
struct pt{
	int x,y;
}p[N];
vector<int> S[N];
cs int tx[5]={0,1,0,-1,0};
cs int ty[5]={-1,0,1,0,0};
int find(int x){
	return fa[x]==x?x:fa[x]=find(fa[x]);
}
inline void merge(int u,int v){
	int f1=find(u),f2=find(v);
	if(f1!=f2)fa[f1]=f2;
}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#else
	#ifndef ONLINE_JUDGE
	#endif
	#endif
	m=read(),n=read();
	tot=n*m;
	for(int i=0;i<m;i++)
	for(int j=0;j<n;j++){
		int x=read();
		a[pos(i,j)]=x;
	}
	q=read();
	for(int i=1;i<=q;i++)fa[i]=i;
	for(int i=1;i<=q;i++){
		int x=read(),y=read(),px,py;
		p[i].x=x,p[i].y=y;
		for(int t=0;t<=4;t++){
			px=x+tx[t],py=y+ty[t];
			if(!(0<=px&&px<m&&0<=py&&py<n))continue;
			if(vis[pos(px,py)])merge(i,vis[pos(px,py)]);
			vis[pos(px,py)]=i;
		}
	}
	for(int i=1;i<=q;i++){
		S[find(i)].pb(i);
	}
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=q;i++)if(S[i].size()){
		int siz=0,mn=1e9,s=0;
		for(int &t:S[i]){
			vis[pos(p[t].x,p[t].y)]=1,s+=a[pos(p[t].x,p[t].y)];
		}
		for(int &t:S[i]){
			int x=p[t].x,y=p[t].y,px,py;
			for(int tt=0;tt<=3;tt++){
				px=x+tx[tt],py=y+ty[tt];
				if(!(0<=px&&px<m&&0<=py&&py<n))continue;
				if(vis[pos(px,py)])continue;
				vis[pos(px,py)]=1,s+=a[pos(px,py)],siz++,chemn(mn,a[pos(px,py)]);
			}
		}
		if(siz<S[i].size()*3){puts("No");return 0;}
		if(siz==S[i].size()*3)ans+=s;
		if(siz>S[i].size()*3)ans+=s-mn;
	}
	cout<<ans;
}