考虑把S设为文科，T设为理科，
S到每个点连边，流量为选文HAPPY+1/2（和四周一起选文的贡献）
T理科同理。
把四周挨着的建双向边条，流量为1/2（两人同选文加同选理的贡献）即可。
读入太扯淡了
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<queue>
#define inf 100000000
using namespace std;
int w_[105][105],l_[105][105],w_h[105][105],w_z[105][105],l_h[105][105],l_z[105][105];
int n,m,S=0,T,cnt,a[105][105],sum,ans;
int adj[10005],e=2,dep[10005];
struct node
{
int v,next,l;
} lu[100004];
int read() {
int s=0,f=1;char ch=getchar();
while(ch>'9'||ch<'0') {if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') {s=(s<<1)+(s<<3)+(ch^48);ch=getchar();}
return s*f;
}
void add(int u,int v,int l)
{lu[e].v=v;lu[e].l=l;lu[e].next=adj[u];adj[u]=e++;}
int bfs()
{
queue<int> q;
memset(dep,0,sizeof(dep));
q.push(S);
dep[S]=1;
while(!q.empty())
{
int x=q.front();q.pop();
for(int i=adj[x];i;i=lu[i].next)
{
if(lu[i].l&&!dep[lu[i].v])
{//cout<<lu[i].l<<" ";
dep[lu[i].v]=dep[x]+1;
q.push(lu[i].v);
if(lu[i].v==T)
return 1;
}
}
}
return 0;
}
int dfs(int x,int len)
{
if(x==T)return len;
int tmp=len,k;
for(int i=adj[x];i;i=lu[i].next)
{
int to=lu[i].v;
if(tmp&&dep[to]==dep[x]+1&&lu[i].l)
{
k=dfs(to,min(lu[i].l,tmp));
if(!k)
{
dep[to]=0;
continue;
}
lu[i].l-=k;
lu[i^1].l+=k;
tmp-=k;
}
}
return len-tmp;
}
int yjn()
{
freopen("nt2011_happiness.in","r",stdin);
freopen("nt2011_happiness.out","w",stdout);
scanf("%d%d",&n,&m);T=n*m+1;
for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)w_ [i][j]=read(),sum+=w_ [i][j];
for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)l_ [i][j]=read(),sum+=l_ [i][j];
for(int i=1;i< n;i++)for(int j=1;j<=m;j++)w_z[i][j]=read(),sum+=w_z[i][j];
for(int i=1;i< n;i++)for(int j=1;j<=m;j++)l_z[i][j]=read(),sum+=l_z[i][j];
for(int i=1;i<=n;i++)for(int j=1;j< m;j++)w_h[i][j]=read(),sum+=w_h[i][j];
for(int i=1;i<=n;i++)for(int j=1;j< m;j++)l_h[i][j]=read(),sum+=l_h[i][j];
for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)a[i][j]=++cnt;
for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
{
add(S,a[i][j],2*w_[i][j]+w_z[i][j]+w_z[i-1][j]+w_h[i][j]+w_h[i][j-1]);add(a[i][j],S,0);
add(a[i][j],T,2*l_[i][j]+l_z[i][j]+l_z[i-1][j]+l_h[i][j]+l_h[i][j-1]);add(T,a[i][j],0);
if(i!=n){add(a[i][j],a[i+1][j],w_z[i][ j ]+l_z[i][ j ]),add(a[i+1][j],a[i][j],w_z[i][ j ]+l_z[i][ j ]);}
if(j!=m){add(a[i][j],a[i][j+1],w_h[i][ j ]+l_h[i][ j ]),add(a[i][j+1],a[i][j],w_h[i][ j ]+l_h[i][ j ]);}
}
while(bfs())
ans+=dfs(S,inf);
cout<<(sum*2-ans)/2;
}
int qty=yjn();
int main(){;}