(poj2559) Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

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.

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

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003

题目大意

   给你几个高度不同，宽度为1的长方形，拼成一张图，求最大矩形面积

题解：

   本题是单调栈(当然你也可以去模拟单调栈或用单调队列来做)

PS：

   关于栈函数的用法介绍，可以看这里(虽然这里说的是队列函数，但是是一样的)

#include 
#include 
#include 
#include 

using namespace std;
int n;
int a[100010],l[100010],r[100010];
stack <int> s;
long long ans;
int main()
{

    while(scanf("%d",&n)&&n)//多组数据
    {
        ans=0;
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        while(!s.empty()) s.pop();
        for(int i=1;i<=n;i++)
        {
            while(!s.empty()&&a[s.top()]>=a[i]) s.pop();
            if(s.empty()) l[i]=0;
                     else l[i]=s.top();
            s.push(i);
        }
        while(!s.empty()) s.pop();
        for(int i=n;i>=1;i--)
        {
            while(!s.empty()&&a[s.top()]>=a[i]) s.pop();
            if(s.empty()) r[i]=n+1;
                     else r[i]=s.top();
            s.push(i);
        }
        for(int i=1;i<=n;i++)
            ans=max(ans,(long long)((long long)r[i]-(long long)l[i]-1LL)*(long long)a[i]);
        printf("%lld\n",ans);
    }
    return 0;
}