题目链接：HDU - 1506
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 that 1 <= 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.
题意描述：给出一些宽度为1，高度大于等于0的矩形，求出最大的矩形面积。
算法分析：看到n的范围，就确定不能n*n了，想了想，对于一个矩形i ，找到最左方和最右连续都比它高的位置之后，对于矩形i+1来说，如果i+1高度比i小，那么对i+1来往左扩展找出左方连续比它高的最左方的位置时候，就没有必要再次比较i+1和i、i+1和i-1、、、因为前面有一部分由 i 已经比较过了，所以我们只需要从上次的记录的位置开始往左比较即可。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cstdlib>
5 #include<cmath>
6 #include<algorithm>
7 #define inf 0x7fffffff
8 using namespace std;
9 typedef long long LL;
10 const int maxn=100000+10;
11
12 int n;
13 int l[maxn],r[maxn],an[maxn];
14
15 int main()
16 {
17 while (scanf("%d",&n)!=EOF && n)
18 {
19 for (int i=1 ;i<=n ;i++) {scanf("%d",&an[i]);l[i]=r[i]=i;}
20 an[0]=an[n+1]=-1;
21 l[0]=l[n+1]=r[0]=r[n+1]=0;
22 for (int i=1 ;i<=n ;i++)
23 {
24 while (an[l[i]-1 ]>=an[i])
25 l[i]=l[l[i]-1 ];
26 }
27 for (int i=n ;i>=1 ;i--)
28 {
29 while (an[r[i]+1 ]>=an[i])
30 r[i]=r[r[i]+1 ];
31 }
32 LL ans=0;
33 for (int i=1 ;i<=n ;i++)
34 {
35 LL area=(LL)(r[i]-l[i]+1)*(LL)an[i];
36 if (area>ans) ans=area;
37 }
38 printf("%I64d\n",ans);
39 }
40 return 0;
41 }