Terrible Sets （单调队列）

最新推荐文章于 2020-03-10 18:35:27 发布

原创最新推荐文章于 2020-03-10 18:35:27 发布 · 322 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#单调队列

单调队列专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种计算多个不同尺寸矩形中最大矩形面积的方法。通过维护两个单调队列来高效地找到每个矩形左侧和右侧的较高矩形数量，从而计算出最大面积。

Problem 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 w1h1+w2h2+...+wnhn < 109.

Output

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

Sample Input

Sample Output

12
14

题目大概：

在x轴上有很多宽度高度不一样的矩形，问最大的矩形面积是多少。

思路：

这个题和前几天做的题都很相似，都是维护两个单调队列，一个求在i这个数左边比这个矩形高的连续矩形有几个，另一个是求右面的，然后就是长宽乘积求最大面积。

代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>

using namespace std;

int n;
int a[55000];
int l[55000];
int r[55000];
int q[55000];
int b[55000];

int getl()
{


    for(int i=2;i<=n;i++)
    {
        while(l[i]>1&&a[i]<=a[l[i]-1])
        l[i]=l[l[i]-1];


    }
}

int getr()
{

      for(int i=n;i>=1;i--)
    {
        while(r[i]<=n&&a[i]<=a[r[i]+1])
        r[i]=r[r[i]+1];


    }
}


int main()
{

    while(scanf("%d",&n))
    {if(n==-1)break;
    memset(a,0,sizeof(a));

    memset(q,0,sizeof(q));

    b[0]=0;
    for(int i=1;i<=n;i++)
    {   int rr;
        scanf("%d%d",&rr,&a[i]);
        b[i]=b[i-1]+rr;
    }


    a[0]=a[n+1]=-1;
    for(int i=1;i<=n;i++)
    {
        l[i]=i;
        r[i]=i;
    }
    getl();
    getr();


   long long sun=-1;
    for(int i=1;i<=n;i++)
    {  long long sum=0;
        sum=(b[r[i]]-b[l[i]-1])*a[i];
        if(sum>sun)sun=sum;
    }
    printf("%d\n",sun);
    }


    return 0;
}