题链：
http://www.lydsy.com/JudgeOnline/problem.php?id=2127
题解：
和 BZOJ 3984 建图类似（最小割模型）。
但是这个建图方法效率有点低。
另外这里这个解法比较高效，写得很好很清晰：
http://blog.****.net/hzj1054689699/article/details/53038620
代码：
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 65000
#define MAXM 300000
#define INF 0x3f3f3f3f
using namespace std;
struct Edge{
int to[MAXM],cap[MAXM],nxt[MAXM],head[MAXN],ent;
void Init(){
ent=2; memset(head,0,sizeof(head));
}
void Adde(int u,int v,int w){
to[ent]=v; cap[ent]=w; nxt[ent]=head[u]; head[u]=ent++;
to[ent]=u; cap[ent]=0; nxt[ent]=head[v]; head[v]=ent++;
}
int Next(int i,bool type){
return type?head[i]:nxt[i];
}
}E;
int cur[MAXN],d[MAXN];
int N,M,S,T,ans;
int idx(int i,int j,int k){
return (i-1)*M+j+k*N*M;
}
bool bfs(){
queue<int>q; int u,v;
memset(d,0,sizeof(d));
d[S]=1; q.push(S);
while(!q.empty()){
u=q.front(); q.pop();
for(int i=E.Next(u,1);i;i=E.Next(i,0)){
v=E.to[i];
if(d[v]||!E.cap[i]) continue;
d[v]=d[u]+1; q.push(v);
}
}
return d[T];
}
int dfs(int u,int reflow){
if(u==T||!reflow) return reflow;
int flowout=0,f,v;
for(int i=E.Next(u,1);i;i=E.Next(i,0)){
v=E.to[i];
if(d[v]!=d[u]+1) continue;
f=dfs(v,min(reflow,E.cap[i]));
flowout+=f; E.cap[i^1]+=f;
reflow-=f; E.cap[i]-=f;
if(!reflow) break;
}
if(!flowout) d[u]=0;
return flowout;
}
int Dinic(){
int flow=0;
while(bfs()){
memcpy(cur,E.head,sizeof(E.head));
flow+=dfs(S,INF);
}
return flow;
}
int main()
{
E.Init();
scanf("%d%d",&N,&M);
S=N*M*5+1; T=N*M*5+2;
for(int i=1,x;i<=N;i++)
for(int j=1;j<=M;j++)
scanf("%d",&x),ans+=x,E.Adde(S,idx(i,j,0),x);
for(int i=1,x;i<=N;i++)
for(int j=1;j<=M;j++)
scanf("%d",&x),ans+=x,E.Adde(idx(i,j,0),T,x);
for(int i=1,_i,_j,x;i<N;i++)
for(int j=1;j<=M;j++){
scanf("%d",&x); ans+=x;
E.Adde(S,idx(i,j,1),x);
E.Adde(idx(i,j,1),idx(i,j,0),INF);
E.Adde(idx(i,j,1),idx(i+1,j,0),INF);
}
for(int i=1,_i,_j,x;i<N;i++)
for(int j=1;j<=M;j++){
scanf("%d",&x); ans+=x;
E.Adde(idx(i,j,2),T,x);
E.Adde(idx(i,j,0),idx(i,j,2),INF);
E.Adde(idx(i+1,j,0),idx(i,j,2),INF);
}
for(int i=1,_i,_j,x;i<=N;i++)
for(int j=1;j<M;j++){
scanf("%d",&x); ans+=x;
E.Adde(S,idx(i,j,3),x);
E.Adde(idx(i,j,3),idx(i,j,0),INF);
E.Adde(idx(i,j,3),idx(i,j+1,0),INF);
}
for(int i=1,_i,_j,x;i<=N;i++)
for(int j=1;j<M;j++){
scanf("%d",&x); ans+=x;
E.Adde(idx(i,j,4),T,x);
E.Adde(idx(i,j,0),idx(i,j,4),INF);
E.Adde(idx(i,j+1,0),idx(i,j,4),INF);
}
int der=Dinic();
ans-=der;
printf("%d",ans);
return 0;
}