【BZOJ3894】文理分科（最小割）

题意：给定一个m*n的矩阵，每个格子的人可以学文或者学理，学文和学理各有一个满意度，

如果以某人为中心的十字内所有人都学文或者学理还会得到一个额外满意度，求最大满意度之和

N,M<=100,读入数据均<=500

思路：RYZ作业

惊奇地发现其他人一年前就A了

每个人选文选理很好连，源集连文，汇集连理，文理连OO

对于十字，将每一个十字看作一个点，以都选文举例

源点连到这个点，容量为额外的满意度

这个点连十字中每个点选文，容量为OO

选理同理

答案为所有的满意度之和减去最小割，即放弃的满意度

花费最小的代价，使源集与汇集之间没有流量

const dx:array[1..4]of longint=(-1,1,0,0);
      dy:array[1..4]of longint=(0,0,-1,1);
var head,vet,next,len,gap,dis,fan:array[0..1100000]of longint;
    a,num:array[1..100,1..100,1..4]of longint;
    n,m,tot,i,j,sum,s,source,src,k,x,y:longint;

procedure add(a,b,c:longint);
begin
 inc(tot);
 next[tot]:=head[a];
 vet[tot]:=b;
 len[tot]:=c;
 head[a]:=tot;

 inc(tot);
 next[tot]:=head[b];
 vet[tot]:=a;
 len[tot]:=0;
 head[b]:=tot;
end;

function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;

function dfs(u,aug:longint):longint;
var e,v,val,flow,t:longint;
begin
 if u=src then exit(aug);
 e:=head[u]; val:=s-1; flow:=0;
 while e<>0 do
 begin
  v:=vet[e];
  if len[e]>0 then
  begin
   if dis[u]=dis[v]+1 then
   begin
    t:=dfs(v,min(len[e],aug-flow));
    len[e]:=len[e]-t;
    len[fan[e]]:=len[fan[e]]+t;
    flow:=flow+t;
    if dis[source]>=s then exit(flow);
    if aug=flow then break;
   end;
   val:=min(val,dis[v]);
  end;
  e:=next[e];
 end;
 if flow=0 then
 begin
  dec(gap[dis[u]]);
  if gap[dis[u]]=0 then dis[source]:=s;
  dis[u]:=val+1;
  inc(gap[dis[u]]);
 end;
 exit(flow);
end;

function maxflow:longint;
var ans:longint;
begin
 fillchar(gap,sizeof(gap),0);
 fillchar(dis,sizeof(dis),0);
 gap[0]:=s; ans:=0;
 while dis[source]<s do ans:=ans+dfs(source,maxlongint);
 exit(ans);
end;

begin
 assign(input,'bzoj3894.in'); reset(input);
 assign(output,'bzoj3894.out'); rewrite(output);
 readln(n,m);
 for i:=1 to 1100000 do
  if i and 1=1 then fan[i]:=i+1
   else fan[i]:=i-1;
 for i:=1 to 4 do
  for j:=1 to n do
   for k:=1 to m do
   begin
    read(a[j,k,i]);
    inc(s); num[j,k,i]:=s;
    sum:=sum+a[j,k,i];
   end;
 source:=s+1; src:=s+2;
 s:=s+2;
 for i:=1 to n do
  for j:=1 to m do add(source,num[i,j,1],a[i,j,1]);
 for i:=1 to n do
  for j:=1 to m do
  begin
   add(source,num[i,j,3],a[i,j,3]);
   for k:=1 to 4 do
   begin
    x:=i+dx[k]; y:=j+dy[k];
    if (x>0)and(x<=n)and(y>0)and(y<=m) then add(num[i,j,3],num[x,y,1],maxlongint);
   end;
   add(num[i,j,3],num[i,j,1],maxlongint);
  end;
 for i:=1 to n do
  for j:=1 to m do add(num[i,j,1],num[i,j,2],maxlongint);
 for i:=1 to n do
  for j:=1 to m do add(num[i,j,2],src,a[i,j,2]);
 for i:=1 to n do
  for j:=1 to m do
  begin
   add(num[i,j,4],src,a[i,j,4]);
   for k:=1 to 4 do
   begin
    x:=i+dx[k]; y:=j+dy[k];
    if (x>0)and(x<=n)and(y>0)and(y<=m) then add(num[x,y,2],num[i,j,4],maxlongint);
   end;
   add(num[i,j,2],num[i,j,4],maxlongint);
  end;
 writeln(sum-maxflow);
end.




 close(input);
 close(output);
end.