单一栈-棋盘制作

program as;
var
   a,b       : array[0..2000,0..2000]of longint;
   c       : array[0..2000,1..2]of longint;
   n,m,i,j,l,cfx,zfx : longint;

function min(a,b : longint):longint;
begin
   if a>b then exit(b);
   exit(a);
end;
procedure chuli(l,r,j:longint);
begin
   for l:=l to r do
   begin
      if c[l,1] *(j-c[l,2])>cfx then
  cfx:=c[l,1] *(j-c[l,2]);
      if ((j-c[l,2])>zfx)and(j-c[l,2]>zfx) then
  zfx:=min(c[l,1],j-c[l,2]);
   end;
end;
begin
   readln(n,m);
   for i:=1 to n do
   begin
      for j:=1 to m do
      begin
  read(a[i,j]);
  if (i=1)or(a[i,j]=a[i-1,j]) then
  begin
     b[i,j]:=1;
  end
  else
  begin
     b[i,j]:=b[i-1,j]+1;
  end;
      end;
      readln;
   end;
   for i:=1 to n do
   begin
      l:=1;
      c[1,1]:=b[i,1];
      c[1,2]:=1;
      for j:=2 to m do
      begin
  if a[i,j]<>a[i,j-1] then
  begin
     while c[l,1]>b[i,j] do
     begin
        chuli(l,l,j);
        if c[l-1,1]<b[i,j] then
        begin
    c[l,1]:=b[i,j];
    break;
        end
        else
        begin
    dec(l);
        end;
     end;
     if c[l,1]<b[i,j] then
     begin
        inc(l);
        c[l,1]:=b[i,j];
        c[l,2]:=j;
     end;
  end
  else
  begin
     chuli(1,l,j);
     l:=1;
     c[l,1]:=b[i,j];
     c[l,2]:=j;
  end;
      end;
      j:=m+1;
      chuli(1,l,j);
   end;
   writeln(zfx*zfx);
   writeln(cfx);
end.

这道题刚做的时候觉得比队列简单可是，做了才知道不好处理（队列要控制两个指针，而栈就控制一个），最应该注意的是出栈时的处理，分为两中出栈一种是全出，另一种是a[n]到a[n-1],a[n]到a[n-2],a[n]到a[n-3].......a[n]到a[1],

最后还要考虑到全出栈的情况。