贴个单调队列的代码 写的有些麻烦了
两个n*n的循环就够了
为了美观  就写了4个循环
program dsqwwe;
  var
    q:array[0..1000] of longint;
    g1,g2,f1,f2,a:array[0..250,0..250] of longint;
    open,closed,i,j,n,b,k,x,y:longint;
  begin
    assign(input,'poj2019.in');
    reset(input);
    assign(output,'poj2019.out');
    rewrite(output);

    readln(n,b,k);
    for i:=1 to n do
     for j:=1 to n do
      read(a[i,j]);
    //max////////////////////////////
    for i:=1 to n do
     begin
      fillchar(q,sizeof(q),0);
      closed:=1;
      open:=1;
      q[1]:=1;
      for j:=1 to n do
       begin
         while (closed<open) and (j-q[closed]>b-1) do inc(closed);
         g1[i,j]:=a[i,q[closed]];
         if j<n then
         while (closed<=open) and (a[i,j+1]>=a[i,q[open]]) do dec(open);
         inc(open);
         q[open]:=j+1;
       end;
     end;
   for j:=1 to n do
    begin
     fillchar(q,sizeof(q),0);
     closed:=1;
     open:=1;
     q[1]:=1;
     for i:=1 to n do
      begin
        while (closed<open) and (i-q[closed]>b-1) do inc(closed);
        f1[i,j]:=g1[q[closed],j];
        if i<n then
        while (closed<=open) and (g1[i+1,j]>=g1[q[open],j]) do dec(open);
        inc(open);
        q[open]:=i+1;
      end;
    end;
   //min////////////////////////////////
   for i:=1 to n do
     begin
      fillchar(q,sizeof(q),0);
      closed:=1;
      open:=1;
      q[1]:=1;
     for j:=1 to n do
       begin
         while (closed<open) and (j-q[closed]>b-1) do inc(closed);
         g2[i,j]:=a[i,q[closed]];
         if j<n then
         while (closed<=open) and (a[i,j+1]<=a[i,q[open]]) do dec(open);
         inc(open);
         q[open]:=j+1;
       end;
     end;
   for j:=1 to n do
    begin
     fillchar(q,sizeof(q),0);
     closed:=1;
     open:=1;
     q[1]:=1;
     for i:=1 to n do
      begin
        while (closed<open) and (i-q[closed]>b-1) do inc(closed);
        f2[i,j]:=g2[q[closed],j];
        if i<n then
        while (closed<=open) and (g2[i+1,j]<=g2[q[open],j]) do dec(open);
        inc(open);
        q[open]:=i+1;
      end;
    end;
  //ans/////////////////////////////////////
  for i:=1 to k do
   begin
     readln(x,y);
     writeln(f1[x+b-1,y+b-1]-f2[x+b-1,y+b-1]);
   end;

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