Sliding Window

最新推荐文章于 2023-09-23 11:06:38 发布

原创最新推荐文章于 2023-09-23 11:06:38 发布 · 390 阅读

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本文介绍了一种使用单调队列解决滑动窗口中最大值和最小值问题的算法。通过实例说明了如何在数组中移动固定大小的窗口，并在每个位置找到窗口内的最大和最小数值。

Description

An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
在这里插入图片描述

Your task is to determine the maximum and minimum values in the sliding window at each position.

Input

The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.
Output

There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.

Sample Input
8 3
1 3 -1 -3 5 3 6 7

Sample Output
-1 -3 -3 -3 3 3
3 3 5 5 6 7

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分析
单调队列即可
注意m=1的情况

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程序：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int a[1000010],q[1000010],n,m,p[1000010];

void zx()
{
	int head=1,tail=1;
	q[head]=1;
	if (m==1) printf("%d ",a[q[head]]);
	for (int i=2;i<=n;i++)
	{
		while (head<=tail&&a[q[tail]]>=a[i]) tail--;
		tail++;
		q[tail]=i;
		while (i-q[head]+1>m) head++;
		if (i>=m) printf("%d ",a[q[head]]);
	}
}

void zd()
{
	int head=1,tail=1;
	p[head]=1;
	if (m==1) printf("%d ",a[p[head]]);
	for (int i=2;i<=n;i++)
	{
		while (head<=tail&&a[p[tail]]<=a[i]) tail--;
		tail++;
		p[tail]=i;
		while (i-p[head]+1>m) head++;
		if (i>=m) printf("%d ",a[p[head]]);
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	zx();
	cout<<endl;
	zd();
	return 0;
}