poj2559 Largest Rectangle in a Histogram（单调栈）

Description

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles: 

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

Input

The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that1<=n<=100000. Then follow n integers h₁,...,h_n, where 0<=h_i<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.

Output

For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.

Sample Input

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003

 

 

题目大意：

这题就是个单调栈。贴一下书上的题解吧：

代码：

 program rrr(input,output);
 var
   q:array[..]of longint;
   a,l,r:array[..]of int64;
   n,i,t:longint;
   ans:int64;
 function max(a,b:int64):int64;
 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
          read(n);if n= then break;
          for i:= to n do read(a[i]);
          t:=;q[]:=;
          for i:= to n do
             begin
                while (t>) and (a[i]<=a[q[t]]) do dec(t);
                l[i]:=q[t];
                inc(t);q[t]:=i;
             end;
          t:=;q[]:=n+;
          for i:=n downto  do
             begin
                while (t>) and (a[i]<=a[q[t]]) do dec(t);
                r[i]:=q[t];
                inc(t);q[t]:=i;
             end;
          ans:=;
          for i:= to n do ans:=max(ans,a[i]*(r[i]-l[i]-));
          writeln(ans);
       end;
    close(input);close(output);
 end.