POJ 2559 Largest Rectangle in a Histogram

最新推荐文章于 2022-01-16 20:26:37 发布

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本文介绍了一种使用单调栈求解直方图中最大矩形面积的问题，通过保持栈顶元素最优的方式，有效地解决了计算最大矩形面积的问题。文章提供了完整的代码实现。

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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

一道单调栈的模板题：
初始化栈为空，并在之后始终保持队首元素为最值
将当前元素和栈顶元素比较，若优于栈顶元素，则栈顶元素出栈直到当前元素不优于栈顶元素，当前元素进栈
对于原数组每一个元素 ai 都执行以上的操作

#include<iostream>
#include<cstring>
#include<cstdio>
#include<stack>
using namespace std;
stack<int>stk;
int n,len[100001],h[100001];
int main()
{
    while(scanf("%d",&n)==1&&n)
    {
        long long ans=0;
        memset(len,0,sizeof(len));
        for(int i=1;i<=n;i++)
            scanf("%d",&h[i]);
        for(int i=1;i<=n;i++)
        {
            while(!stk.empty()&&h[i]<h[stk.top()])
            {
                len[i]+=len[stk.top()];
                ans=max(ans,(long long)len[i]*h[stk.top()]);
                stk.pop();
            }
            len[i]++;
            stk.push(i);
        }
        int l=0;
        while(!stk.empty())
        {
            l+=len[stk.top()];
            ans=max(ans,(long long)h[stk.top()]*l);
            stk.pop();
        }
        printf("%lld\n",ans);
    }
    return 0;
}