Largest Rectangle in a Histogram

题目描述

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.

这道题让求直方图中最大的矩形

输入和输出

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.

输入包含多组数据。每组数据包含一行，第一个正整数N表示有N个矩形，接下来N个正整数描述其高度。 数据以N=0结束。 数据范围参看英文题面。

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.

对于每组数据输出一行包含一个正整数，表示该组数据的答案。

样例

Sample Input

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

数据规模与约定

时间限制：1s

空间限制：64MB

　　这道题刚开始我的想法是依次枚举通过向两边不断延展，得到矩形的宽，求得面积，依次比较，得出答案。这种思路没问题，但是时间是绝对超时的，经过一番激烈的思想斗争，在y总的指导下也是A了这道题。正确思路如下;

 

　　在原有的思路上我们加一个单调栈q，将不符合（低于楼层高度）的不断弹出栈，记录左右两端所扩展的范围，算出矩阵面积，比较即可。（在acwing上y总的最后求宽时我认为是手误，多加减了一个一把）。

代码如下：

#include<bits/stdc++.h>
using namespace std;

int n;
int q[100030],r[100030],l[100030],h[100030];

void work(int a[100030])
{
　　int tot=0;
　　h[0]=-1;
　　for(int i=1;i<=n;i++)
　　{
　　　　while(h[q[tot]]>=h[i]) tot--;
　　　　a[i]=q[tot];
　　　　q[++tot]=i;
　　}
}

int main()
{
　　while(cin>>n,n)
　　{
　　　　for(int i=1;i<=n;i++) scanf("%d",&h[i]);
　　　　work(l);
　　　　reverse(h+1,h+n+1);//翻转省步骤
　　　　work(r);
　　　　long long ans=0;
　　　　for(int i=1,j=n;i<=n;i++,j--)
　　　　{
　　　　　　long long side;
　　　　　　side=n-l[j]-r[i];
　　　　　　ans=max(ans,h[i]*side);
　　　　}　　
　　　　cout<<ans<<endl;
　　}
　　return 0;
}

❀完结撒花❀