Terrible Sets --- 单调栈解决问题

最新推荐文章于 2020-03-10 18:35:27 发布

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本文介绍了一种求解由给定数据集构成的最大矩形面积的方法，通过使用栈来保持高度的单调递增顺序，并逐步计算每个可能的矩形面积以找出最大值。

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Description

Let N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N, and w0 = 0.
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑ _0<=j<=i-1wj <= x <= ∑ _0<=j<=iwj}
Again, define set S = {A| A = WH for some W , H ∈ R ⁺ and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}.
Your mission now. What is Max(S)?
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy.
But for this one, believe me, it's difficult.

Input

The input consists of several test cases. For each case, n is given in a single line, and then followed by n lines, each containing wi and hi separated by a single space. The last line of the input is an single integer -1, indicating the end of input. You may assume that 1 <= n <= 50000 and w ₁h ₁+w ₂h ₂+...+w _nh _n < 10 ⁹.

Output

Simply output Max(S) in a single line for each case.

Sample Input

Sample Output

12
14

此题就是为了了求解最大的矩形的面积

#include<iostream>

#include<cstdio>

#include<cstring>
#include<algorithm>
#include<stack>
using namespace std;
const int MAXN=100000+10;

int n;

struct node{
	int w,h;
}p[MAXN];
stack<node>s;
int main()
{
	while(scanf("%d",&n)!=EOF&&n!=-1){
		for(int i=1;i<=n;i++)
			scanf("%d%d",&p[i].w,&p[i].h);
		int ans=0;
		for(int i=1;i<=n;i++){
			if(s.empty())
				s.push(p[i]);
			else{
				node temp=s.top();
				if(p[i].h>temp.h)            //保证这个栈是一个按照高进行单调递增的否则 进行替换   然后删除以前的元素；
					s.push(p[i]);
				else{
					int len=0;
					while(!s.empty() && s.top().h>=p[i].h){
						temp=s.top();
						len+=temp.w;
						ans=max(ans,len*temp.h);
						s.pop();
					}
					temp.w=len+p[i].w;   //更新栈顶元素；
					temp.h=p[i].h;         
					s.push(temp);
				}
			}
		}
		int len=0;
		while(!s.empty()){
			node temp=s.top(); s.pop();
			len+=temp.w;
			ans=max(ans,len*temp.h);
		}
		printf("%d\n",ans);
	}
	return 0;
}