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NC50965. Largest Rectangle in a Histogram

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.

输入描述

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 . Then follow n integers , where . 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.

输出描述

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.

示例1

输入:

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

输出:

8
4000

说明:

Huge input, scanf is recommended.

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++(g++ 7.5.0) 解法, 执行用时: 33ms, 内存消耗: 1584K, 提交时间: 2023-07-24 10:40:42

#include<bits/stdc++.h>
using namespace std;
long long a[1000001],p,w[1000001],s[1000001];
int main(){
	long long n,ans;
	while(scanf("%lld",&n)&&n){
		ans=0;
		for(register int i=1;i<=n;++i)scanf("%lld",&a[i]);
		a[n+1]=p=0;
		for(register int i=1;i<=n+1;++i){
			if(a[i]>s[p])s[++p]=a[i],w[p]=1;
			else{
				long long width=0;
				while(s[p]>a[i])width+=w[p],ans=max(ans,width*s[p]),--p;
				s[++p]=a[i],w[p]=width+1;
			}
		}
		printf("%lld\n",ans);
	}
}

C++(clang++ 11.0.1) 解法, 执行用时: 99ms, 内存消耗: 1172K, 提交时间: 2022-08-07 18:52:57

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e5+10;
ll h[N];
int main()
{
	int n;
	while(cin>>n){
		if(n==0)break;
		stack<int>st;
		for(int i=0;i<n;i++)cin>>h[i];
		ll ans=0;
		h[n]=0;
		for(int i=0;i<=n;i++){
			while(!st.empty()&&h[st.top()]>=h[i]){
				int t=st.top();
				st.pop();
				ans=max(ans,h[t]*(st.empty()?i:(i-st.top()-1)));
			}
			st.push(i);
		}
		cout<<ans<<endl;
	}
	
	return 0;
}

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