POJ 2559 Largest Rectangle in a Histogram（单调栈）@

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,...,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.

给定N个宽度相等的高度给定的长方形，从左到右依次排列，求最大的长方形面积；

因为最大的长方形一定是一个最小的长方形的高，能延伸到左边和右边的界限的长方形一定是比他低，如果比他高的话就还能延伸，与假设矛盾。利用单调栈实现

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int N = 1e5+10;
const int mod = 1e9+7;
vectorG[N];
int a[N], st[N], l[N], r[N];


int main()
{
    int n;
    while(scanf("%d", &n),n!=0)
    {
        for(int i=0;i0&&a[st[t-1]]>=a[i]) t--;
            l[i]=t==0?0:st[t-1]+1;
            st[t++]=i;
        }
        t=0;
        for(int i=n-1;i>=0;i--)
        {
            while(t>0&&a[st[t-1]]>=a[i]) t--;
            r[i]=t==0?n:st[t-1];
            st[t++]=i;
        }
        LL ans=0;
        for(int i=0;i