?其實也就卡了卡常，，，
先考慮沒有same_art和same_science 。
起點用art的流量連向每個點，該點再用science的流量連向終點，斷開哪邊相當于少了哪邊收益。
先全部收益加起來，再減去最小割即可。
那有same這些情況怎么辦呢。
考慮新建節點v,起點以same_art連向v，斷開即代表不獲得這段收益。如何避免這個點不斷開s的同時它覆蓋區的點斷開了s？再用v點向覆蓋區連權值為inf的邊，即可保證覆蓋區的點要斷則該點必斷。same_science同理，在下不加以贅述。
528ms，勉強能卡過，，
#include<bits/stdc++.h> #include<cstdio> #include<cmath> #include<cstring> #include<cstdlib> #include<algorithm> #include<queue> #include<deque> #include<list> #include<set> #include<vector> #include<iostream> #define ll int #define re register #define inf 0x7f7f7ff #define inl inline #define sqr(x) (x*x) //#define eps 1e-8 #define debug printf("debug\n"); //#pragma comment(linker, "/STACK:1024000000,1024000000") //#pragma GCC optimize (2) //#pragma G++ optimize (2) using namespace std; const ll mod=51123987; const ll MAXN=105; inl ll read() {re ll x = 0; re int f = 1;char ch = getchar();while(ch<'0'||ch>'9') { if(ch== '-' ) f = -1; ch = getchar(); }while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x * f; } inl char readc() {char ch=getchar();while(('z'<ch||ch<'a')&&('Z'<ch||ch<'A')) ch=getchar();return ch; } inl void write(re ll x){if(x>=10)write(x/10);putchar(x%10+'0'); } inl void writeln(re ll x){if(x<0) {x=-x;putchar('-');}write(x); puts(""); } inl ll gcd(re ll x,re ll y){while(y^=x^=y^=x%=y);return x;} inl ll Lcm(re ll a,re ll b) {return a/gcd(a,b)*b;} inl void FR() {freopen(".in","r",stdin);freopen(".out","w",stdout); } inl void FC() {fclose(stdin);fclose(stdout); } struct edge{ll u,v,w,nxt; }e[400005<<1]; ll ans,n,m,cnt=1,head[200005<<1],s,t,sst; ll dx[10]={0,0,0,1,-1}; ll dy[10]={0,1,-1,0,0}; inl void adde(ll u,ll v,ll w) {e[++cnt].u=u;e[cnt].v=v;e[cnt].w=w;e[cnt].nxt=head[u];head[u]=cnt; } ll d[200005<<1],cur[200005<<1]; inl bool bfs() {queue<ll>Q;for(re ll i=s;i<=sst;i++) d[i]=0;Q.push(s);d[s]=1;while(!Q.empty()) {re ll x=Q.front();Q.pop();for(re ll h=head[x];h;h=e[h].nxt) {re ll v=e[h].v;if(e[h].w&&!d[v]) {d[v]=d[x]+1;if(v==t) return true ;Q.push(v);}}}return d[t]; } ll dfs(ll x,ll limit) {if(x==t) return limit;ll used=0;for(re ll h=cur[x];h;h=e[h].nxt) {if(used==limit) break;if(d[e[h].v]==d[x]+1&&e[h].w) {ll t=dfs(e[h].v,min(limit-used,e[h].w));used+=t;e[h].w-=t;e[h^1].w+=t;}}if(!used) d[x]=0;return used; } int main() { // FR();n=read(),m=read();s=0;t=n*m+1;sst=t;for(re ll i=1;i<=n;i++) {for(re ll j=1;j<=m;j++) {re ll x=read();adde(s,(i-1)*m+j,x);adde((i-1)*m+j,s,0);ans+=x;}}for(re ll i=1;i<=n;i++) {for(re ll j=1;j<=m;j++) {re ll x=read();adde((i-1)*m+j,t,x);adde(t,(i-1)*m+j,0);ans+=x;}}for(re ll i=1;i<=n;i++) {for(re ll j=1;j<=m;j++) {re ll x=read();ans+=x;sst++;adde(s,sst,x);adde(sst,s,0);for(re ll k=0;k<=4;k++) {re ll xx=i+dx[k],yy=j+dy[k];if(xx&&xx<=n&&yy&&yy<=m) {adde(sst,(xx-1)*m+yy,inf);adde((xx-1)*m+yy,sst,0);}}}}for(re ll i=1;i<=n;i++) {for(re ll j=1;j<=m;j++) {re ll x=read();ans+=x;sst++;adde(sst,t,x);adde(t,sst,0);for(re ll k=0;k<=4;k++) {re ll xx=i+dx[k],yy=j+dy[k];if(xx&&xx<=n&&yy&&yy<=m) {adde((xx-1)*m+yy,sst,inf);adde(sst,(xx-1)*m+yy,0);}}}}while(bfs()) {memcpy(cur,head,sizeof(cur));ans-=dfs(s,inf);}writeln(ans); // FC();return 0; }
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