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;
}