Largest Rectangle in a Histogram-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_61362165/article/details/128984646

文章描述了一种用于计算直方图中最大矩形面积的算法，该算法使用栈来处理矩形高度，并通过比较和更新找到最大面积。输入包含多个测试用例，每个用例有若干矩形的高度，程序需找出与基线对齐的最大矩形面积。代码示例使用C++实现，通过扫描矩形高度并维护一个栈来跟踪最大宽度。

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题目：

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

Inputcopy	Outputcopy
7 2 1 4 5 1 3 3 4 1000 1000 1000 1000 0	8 4000

Hint

Huge input, scanf is recommended.

思路：

对于每一组数据，用数组栈a，存储直方图矩形的高度，b[top]表示栈顶的矩形入栈时，a[top]高度且与该矩形相接的最大矩形宽度。

若h比栈顶元素高，则入栈，h高度且与该矩形相接的最大矩形宽度为1。

若h比栈顶元素小或等于，则使高于或等于h的元素出栈，并更新答案，和记录h高度且与该矩形相接的最大矩形宽度。当前栈顶元素为高度的且与该矩形相接的最大矩形宽度为：所有出栈元素的b[i]相加，在加上自身宽度1

代码：

#include<stdio.h>
#include<string.h> 
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1e5 + 5;
ll a[N],b[N],h; 
int n;
int main()
{
	while (scanf("%d",&n) * n)
	{
		memset(a,-1,sizeof(a));
		int t = 0;
		ll ans  = 0;
		for (int i = 1; i <= n + 1; i++){
			if (i != n + 1)	{
				scanf("%lld",&h);
			}
			else h = 0; 
			ll cnt = 0;
			while (h <= a[t]){
				ans = max(ans,(cnt + b[t]) * a[t]);
				cnt += b[t--];
			}
			a[++t] = h;//h元素入栈 
			b[t] = cnt + 1;
		}
		printf("%lld\n",ans);
	}
	return 0;
}