Largest Rectangle in a Histogram POJ - 2559 单调栈

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#POJ #单调栈

博客围绕直方图中最大矩形面积计算展开，介绍了直方图由等宽但高度可能不同的矩形组成，给出输入包含测试用例、矩形数量及高度等信息，输出为每个测试用例中指定直方图里最大矩形的面积，还给出了输入输出示例及提示。

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.

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
int n, top;
const int maxn = 107777;
ll num[maxn], l[maxn], r[maxn], sta[maxn];
int main()
{
    while(scanf("%d", &n) != EOF && n)
    {
       top = 0;
       for(int i = 0; i < n; i++)
           scanf("%lld", &num[i]);
       l[0] = 0;
       sta[top++] = 0;
       for(int i = 1; i < n; i++)
       {
           if(num[i] > num[sta[top - 1]])
           {
               l[i] = sta[top - 1] + 1;
               sta[top++] = i;
           }
           else
           {
               while(top > 0 && num[i] <= num[sta[top - 1]])
                top--;
               if(top == 0) l[i] = 0;
               else l[i] = sta[top - 1] + 1;
               sta[top++] = i;
           }
       }
       top = 0;
       r[n - 1] = n - 1;
       sta[top++] = n - 1;
       for(int i = n - 2; i >= 0; i--)
       {
           if(num[i] > num[sta[top - 1]])
           {
               r[i] = i;
               sta[top++] = i;
           }
           else
           {
               while(top > 0 && num[i] <= num[sta[top - 1]])
                top--;
               if(top == 0)r[i] = n - 1;
               else r[i] = sta[top - 1] - 1;
               sta[top++] = i;
           }
       }
       ll res = 0;
       for(int i = 0; i < n; i++)
       {
           res = max(res,(r[i] - l[i] + 1) * num[i]);
       }
       cout<<res<<endl;

    }
    return 0;
}