【Codevs1907】方格取数3（最小割）

题意：在一个有m*n 个方格的棋盘中，每个方格中有一个正整数。现要从方格中取数，使任
意2 个数所在方格没有公共边，且取出的数的总和最大。试设计一个满足要求的取数算法。

n,m<=30

思路：如果将棋盘黑白点染色，可以发现相邻的黑白点不能同时取

将源点到黑点连一条容量为黑点数字的边，黑点到相邻白点连容量为∞的边，白点到汇点连容量为白点数字的边，可以发现如果不让相邻黑白点同时取到，三条边中必定要切断一条

题意就是让切断的总和最小，所以显然不会切∞，只会切另外两条

跑最小割即可

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

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

 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]:=;
  head[b]:=tot;
 end;

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

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

 begin
  assign(input,'Codevs1907.in'); reset(input);
  assign(output,'Codevs1907.out'); rewrite(output);
  readln(n,m);
  for i:= to  do
   if i mod = then fan[i]:=i+
    else fan[i]:=i-;
  for i:= to n do
   for j:= to m do
   begin
    read(a[i,j]); sum:=sum+a[i,j];
    inc(s); num[i,j]:=s; b[i,j]:=(i+j+) mod ;
   end;
  for i:= to n do
   for j:= to m do
    if b[i,j]= then
     for k:= to  do
     begin
      x:=i+dx[k]; y:=j+dy[k];
      if (x>)and(x<=n)and(y>)and(y<=m) then
      begin
       add(num[i,j],num[x,y],maxlongint);
       //writeln(i,' ',j,' ',x,' ',y);
      end;
     end;
  source:=s+; src:=s+; s:=s+;
  for i:= to n do
   for j:= to m do
    if b[i,j]= then
    begin
     //writeln(source,' ',i,' ',j);
     add(source,num[i,j],a[i,j]);
    end
     else
     begin
      //writeln(i,' ',j,' ',src);
      add(num[i,j],src,a[i,j]);
     end;
  writeln(sum-maxflow);

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