BZOJ1001 狼抓兔子 题解

裸的最小割，转化成最大流即可。

#include <bits/stdc++.h>
int n,m;
int S,T;
int mincost;
int head[6001000],tot=1;
int cur[6001000];
int deep[1001000];
int vis[1001000];
std::queue<int>q;
struct qwq{
	int to;
	int nxt;
	int flow;
}e[6001000];
int num(int x,int y){
	return (x-1)*m+y;
}
void add(int x,int y,int z){
  e[++tot]={y,head[x],z};
  head[x]=tot;
  e[++tot]={x,head[y],z};
  head[y]=tot;
}
bool bfs(){
	memset(deep,0,sizeof deep);
  deep[S]=1;
  q.push(S);
  while(!q.empty()){
  	int X=q.front();
  	q.pop();
  	for(int i=head[X];i;i=e[i].nxt){
  		int y=e[i].to;
  		if(!deep[y]&&e[i].flow){
  			deep[y]=deep[X]+1;
  			q.push(y);
  		}
  	}
  }
  return deep[T];
}
int dfs(int x,int flow){
	if(x==T||!flow)
	  return flow;
	int Flow=0;
	for(int &i=cur[x];i;i=e[i].nxt){
		int y=e[i].to;
		if(e[i].flow&&deep[y]==deep[x]+1){
			int w=dfs(y,std::min(flow,e[i].flow));
			if(w){
				e[i].flow-=w;
				e[i^1].flow+=w;
				Flow+=w;
				flow-=w;
				if(!flow)
				  break;
			}
		}
	}
	return Flow;
}
void dinic(){
	int maxflow=0;
	while(bfs()){
		memcpy(cur,head,sizeof head);
		while(1){
			int w=dfs(S,0x3f3f3f3f);
			if(!w)
			  break;
			maxflow+=w;
		}
	}
	printf("%d\n",maxflow);
}
main(){
	scanf("%d%d",&n,&m);
	S=num(1,1);
	T=num(n,m);
	for(int i=1;i<=n;++i)
	  for(int j=2;j<=m;++j){
	  	int x;
	  	scanf("%d",&x);
	    add(num(i,j-1),num(i,j),x);
	  }
	for(int i=2;i<=n;++i)
	  for(int j=1;j<=m;++j){
	  	int x;
	  	scanf("%d",&x);
	    add(num(i-1,j),num(i,j),x);
	  }
	for(int i=2;i<=n;++i)
	  for(int j=2;j<=m;++j){
	  	int x;
	  	scanf("%d",&x);
	    add(num(i-1,j-1),num(i,j),x);
	  }
	dinic();
	return 0;
}