【BZOJ1412】狼和羊的故事（最小割）

题意：将一个由0,1,2构成的矩阵里的1与2全部分割最少需要选取多少条边

n,m<=100

思路：裸的最小割模型

相邻的格子连容量为1的边（其实可以少连很多遍，1与1，2与2之间的边是没有意义的）

由源点到所有1连容量为oo的边，2到汇点连容量为oo的边

最小割即是答案

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

 procedure add(a,b,c:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  len[tot]:=c;
  head[a]:=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,t,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(aug-flow,len[e]));
     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,'bzoj1412.in'); reset(input);
  assign(output,'bzoj1412.out'); rewrite(output);
  readln(n,m);
  for i:= to n do
   for j:= to m do
   begin
    read(a[i,j]);
    inc(s); num[i,j]:=s;
   end;
  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
    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],);
      add(num[x,y],num[i,j],);
     end;
    end;
  source:=s+; src:=s+; s:=s+;
  for i:= to n do
   for j:= to m do
   begin
    if a[i,j]= then
    begin
     add(source,num[i,j],maxlongint);
     add(num[i,j],source,);
    end;
    if a[i,j]= then
    begin
     add(num[i,j],src,maxlongint);
     add(src,num[i,j],);
    end;
   end;

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