poj2082 Terrible Sets（单调栈）

Description

Let N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N, and w0 = 0. 
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑_0<=j<=i-1wj <= x <= ∑_0<=j<=iwj} 
Again, define set S = {A| A = WH for some W , H ∈ R⁺ and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}. 
Your mission now. What is Max(S)? 
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy. 
But for this one, believe me, it's difficult.

Input

The input consists of several test cases. For each case, n is given in a single line, and then followed by n lines, each containing wi and hi separated by a single space. The last line of the input is an single integer -1, indicating the end of input. You may assume that 1 <= n <= 50000 and w₁h₁+w₂h₂+...+w_nh_n < 10⁹.

Output

Simply output Max(S) in a single line for each case.

Sample Input

3
1 2
3 4
1 2
3
3 4
1 2
3 4
-1

Sample Output

12
14

Source

Shanghai 2004 Preliminary

 

 

刚做了poj2559，这题只是每个矩形有宽度了。

poj2559的题解：http://www.cnblogs.com/Currier/p/6686478.html

program rrr(input,output);
var
  h,l,r,q,sum:array[0..50050]of longint;
  n,i,t,ans,w:longint;
function max(a,b:longint):longint;
begin
   if a>b then exit(a) else exit(b);
end;
begin
   assign(input,'r.in');assign(output,'r.out');reset(input);rewrite(output);
   while true do
      begin
         readln(n);if n=-1 then break;
         sum[0]:=0;
         for i:=1 to n do begin readln(w,h[i]);sum[i]:=sum[i-1]+w; end;
         t:=0;q[0]:=0;
         for i:=1 to n do
            begin
               while (t>0) and (h[i]<=h[q[t]]) do dec(t);
               l[i]:=q[t];
               inc(t);q[t]:=i;
            end;
         t:=0;q[0]:=n+1;
         for i:=n downto 1 do
            begin
               while (t>0) and (h[i]<=h[q[t]]) do dec(t);
               r[i]:=q[t];
               inc(t);q[t]:=i;
            end;
         ans:=0;
         for i:=1 to n do ans:=max(ans,h[i]*(sum[r[i]-1]-sum[l[i]]));
         writeln(ans);
      end;
   close(input);close(output);
end.