I - Largest Rectangle in a Histogram

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题目：

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.

Sample Input

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

Sample Output

8
4000

题意：

让你找出最大的矩形的面积；

思路：

这是一道dp，找出每一个矩形左面和右面的连续比这一个矩形高的矩形的个数，然后记录下来，然后再计算面积，最后再进行计算面积，然后比较找出最大值；

代码如下：

#include<stdio.h>
using namespace std;const int N=100010;
long long int a[N],l[N],r[N],maxn,s,t;int main()
{int n;while(~scanf("%d",&n)&&n){int i,j;for(int i=1; i<=n; i++)scanf("%I64d",&a[i]);l[1]=1;//对于每一个a[i],用dp找出a[i]左边和右边连续大于自己的数的长度r[n]=n;for(i=2; i<=n; i++)//l[i]表示比a[i]大的数连续的最左边的位置{t=i;while(t>1&&a[i]<=a[t-1])t=l[t-1];l[i]=t;}for(i=n-1; i>=1; i--)//r[i]表示比a[i]大的数连续的最右边的位置{t=i;while(t<n&&a[i]<=a[t+1])t=r[t+1];r[i]=t;}maxn=0;for(i=1; i<=n; i++){s=(r[i]-l[i]+1)*a[i];if(s>maxn)maxn=s;}printf("%lld\n",maxn);}return 0;
}

 

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