正解:网络流+对偶图
解题报告:
$umm$日常看不懂题系列了$kk$.其实就是说,给定一个$n\cdot n$的网格图,求最小割$QwQ$
然后网格图的话显然是个平面图,又看到数据范围$n\leq 1000$,显然就考虑平面图转对偶图呗
然后好像就没有什么细节了,,,?
对了,$bzoj$上的话要特判1,洛谷上没有这个数据就不用辣$QwQ$
$QwQ$
(在$bzoj$上$T$了,,,应该是常数的问题懒得改了$QAQ$
#include<bits/stdc++.h>
using namespace std;
#define il inline
//#define int long long
#define fi first
#define sc second
#define gc getchar()
#define mp make_pair
#define P pair<int,int>
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];i;i=edge[i].nxt)
const int N=1500+10;
int n,m,ed_cnt,head[N*N],S,T,dis[N*N];
struct ed{int to,nxt,wei;}edge[N*N*20];
bool vis[N*N];
il int read()
{
rc ch=gc;ri x=0;rb y=1;
while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
if(ch=='-')ch=gc,y=0;
while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
return y?x:-x;
}
il void ad(ri x,ri y,ri z){/*printf("%d -> %d : %d\n",y,x,z);*/edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;}
il int nam(ri x,ri y,ri z){if(!x)return T;if(y>m)return T;if(!y)return S;if(x>n)return S;return (x*m-m+y)*2+z;}
il void dij()
{
priority_queue< P,vector<P>,greater<P> >Q;Q.push(mp(0,S));memset(dis,63,sizeof(dis));
while(!Q.empty())
{
ri nw=Q.top().sc,diss=Q.top().fi;Q.pop();if(vis[nw])continue;vis[nw]=1;dis[nw]=diss;
e(i,nw)if(!vis[t(i)])Q.push(mp(dis[nw]+w(i),t(i)));
}
}
int main()
{
//freopen("4001.in","r",stdin);freopen("4001.out","w",stdout);
n=read()-1;m=read()-1;S=0;T=n*m*2+1;
rp(i,1,n+1){rp(j,1,m){ri tmp=read(),pos1=nam(i,j,-1),pos2=nam(i-1,j,0);/*printf("%d ",tmp);*/ad(pos1,pos2,tmp);ad(pos2,pos1,tmp);}/*printf("\n");*/}
//printf("---\n");
rp(i,1,n){rp(j,1,m+1){ri tmp=read(),pos1=nam(i,j,0),pos2=nam(i,j-1,-1);/*printf("%d ",tmp);*/ad(pos1,pos2,tmp);ad(pos2,pos1,tmp);}/*printf("\n");*/}
//printf("---\n");
rp(i,1,n){rp(j,1,m){ri tmp=read(),pos1=nam(i,j,-1),pos2=nam(i,j,0);/*printf("%d ",tmp);*/ad(pos1,pos2,tmp);ad(pos2,pos1,tmp);}/*printf("\n");*/}
//printf("---\n");
//printf("---------\n");
//rp(i,1,n-1)rp(j,1,m-1)printf("%d %d (%d,%d)\n",nam(i,j,0),nam(i,j,-1),i,j);
dij();printf("%d\n",dis[T]);
return 0;
}
//if(m==1 || n==1) tepan!!!