Largest Rectangle in a Histogram

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：给出了n个矩形的高度，每个矩形宽度为1，花在条形图中，求组成的矩形的最大面积，看样例即可明白。

思路：就是求一下每个矩形左边连续比它大的最小下标计为l[i]，和右边连续比它大的最大下标计为r[i]，然后相减乘以高度即可,最后for循环求出最大面积。代码如下：

#include
#include
long long a[100010];
int l[100010],r[100010];
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        l[1]=1,r[n]=n;
        for(int i=2;i<=n;i++)
        {
            int t=i;
            while(t>1&&a[t-1]>=a[i])
                t=l[t-1]; // 每一个a【i】对应一个l【i】和r【i】。  l[i]是求左边连续比他大的下标最小值

            l[i]=t;
        }
        for(int i=n-1;i>=1;i--)
        {
            int t=i;
            while(t=a[i])
                t=r[t+1];  //r【i】 是求右边连续比他大的下标的最大值
            r[i]=t;   //记录 i矩形的最右端
        }
        long long maxx=0;
        for(int i=1;i<=n;i++)
        {
            long long sum=(r[i]-l[i]+1)*a[i];
            if(sum>maxx)      // 更新maxx的值
                maxx=sum;
        }
        printf("%lld\n",maxx);
    }
}