#include<iostream>
#include<algorithm>
#include<queue>
#include<memory.h>
#include<string>
using namespace std;
const int N=100005,M=360010,INF=14514191;
int S=100001,T=100002;
int h[N],e[M],f[M],ne[M],idx;
int d[N],cur[N];
int n,m,k;
int id(int x,int y){
	if(n>m)return (x-1)*n+y;
	else return x+(y-1)*m;
}
void add(int a,int b,int c){
	e[idx]=b,f[idx]=c,ne[idx]=h[a],h[a]=idx++;
	e[idx]=a,f[idx]=0,ne[idx]=h[b],h[b]=idx++;
}
bool bfs(){
	memset(d,-1,sizeof d);
	queue<int>q;
	q.push(S);
	d[S]=0;cur[S]=h[S];
	while(!q.empty()){
		int t=q.front();q.pop();
		for(int i=h[t];~i;i=ne[i]){
			int to=e[i];
			if(d[to]==-1&&f[i]){
				d[to]=d[t]+1;
				cur[to]=h[to];
				if(to==T)return 1;
				q.push(to);
			}
		}
	}
	return 0;
}
int find(int u,int limit){
    if(u==T)return limit;
    int flow=0;
    for (int i=cur[u];~i&&flow<limit;i=ne[i]){
        cur[u]=i;
        int to=e[i];
        if(d[to]==d[u]+1&&f[i]){
            int t=find(to,min(f[i],limit-flow));
            if(!t)d[to]=-1;
            f[i]-=t,f[i^1]+=t,flow+=t;
        }
    }
    return flow;
}

int dinic(){
	int r=0,flow;
	while(bfs()){
		cout<<"ok"<<endl;
		while(flow=find(S,INF))r+=flow;
	}
	return r;
}
void work(){
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			int v;
			scanf("%d",&v);
			add(id(i,j)+n*m,id(i,j),v);
			//虚点入边 
			if(i>1)add(id(i-1,j),id(i,j)+n*m,INF);
			if(j>1)add(id(i,j-1),id(i,j)+n*m,INF);
			if(i<n)add(id(i+1,j),id(i,j)+n*m,INF);
			if(j<m)add(id(i,j+1),id(i,j)+n*m,INF);
			//实点出边
			if(i>1)add(id(i,j),id(i-1,j),INF);
			if(j>1)add(id(i,j),id(i,j-1),INF);
			if(i<n)add(id(i,j),id(i+1,j),INF);
			if(j<m)add(id(i,j),id(i,j+1),INF);
		}
	}
	for(int i=1;i<=k;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		add(S,id(x,y)+n*m,INF);
	}
	for(int i=1;i<=k;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		add(id(x,y),T,INF);
	}
	printf("%d\n",dinic());
}
signed main(){
	work();
}

在出网络流题，结果发现std死循环。

具体题面：入点连 $k$ 个节点，另外 $k$ 个节点连出点，不能让入点连到出点，删点的代价为 $v_{i,j}$，求最小代价。