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.
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,...,hn, where 0<=hi<=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.