【USACO】Big Barn,1997 Fall

题目本身不难 但是要用滚动数组，否则10000*10000的integer会爆128M

考虑记sqrr[i,j] 为以[i,j]为右下角向左上扩展的正方形的最大值，up[i,j]为以[i,j]为最后一格能向上扩展的最大值,left[i,j]同理

则sqrr[i,j]=min(sqrr[i-1,j-1]+1,up[i,j],left[i,j]);

可以看到第i行状态只与i-1行有关

【NOTICE】 j要downto,否则会覆盖前一行数据

边界：对第一行，如果a[i,j]可行则(sqrr[j]=1,left[j]=left[j-1]+1,up[j]=1)否则均为0

由于算min值要用到当前行的up,left，所以处理每一行前先更新up和left,然后j递减来推sqrr

var
   n,m,i,j,x,y,max:integer;
   a:array[0..10001,0..10001] of boolean;
   up,sqrr,left:{array[0..10001,0..10001] of integer;} array[0..10001] of integer;
function min(a,b,c:integer):integer;
begin
   if (a<=b)and(a<=c) then exit(a);
   if (b<=c)and(b<=a) then exit(b);
   exit(c);
end;
begin
   assign(input,'bigbrn.in');
   reset(input);
   assign(output,'bigbrn.out');
   rewrite(output);
   readln(n,m);
   fillchar(a,sizeof(a),true);
   fillchar(up,sizeof(up),0);
   for i:=1 to m do
   begin
      readln(x,y);
      a[x,y]:=false;
   end;

   for i:=1 to n do if a[1,i] then
   begin
      left[i]:=left[i-1]+1;
      sqrr[i]:=1;
      up[i]:=1;
      if sqrr[i]>max then max:=sqrr[i];
   end;

   for i:=2 to n do
   begin

      for j:=1 to n do
      if a[i,j] then
      begin
        inc(up[j]);
        left[j]:=left[j-1]+1;
      end else
      begin
         up[j]:=0;left[j]:=0;
      end;

      for j:=n downto 2 do
      if a[i,j]  then
      begin
         sqrr[j]:=min(sqrr[j-1]+1,left[j],up[j]);
         if sqrr[j]>max then max:=sqrr[j];

      end   else sqrr[j]:=0;
      if a[i,1] then sqrr[1]:=1 else sqrr[1]:=0;
   end;

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