POJ 2823 Sliding Window 单调队列

最新推荐文章于 2021-01-25 20:02:52 发布

窦小雨

最新推荐文章于 2021-01-25 20:02:52 发布

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分类专栏：数据结构—单调队列文章标签： poj 单调队列

本文链接：https://blog.youkuaiyun.com/qq_33402621/article/details/52458635

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本文介绍了一种利用单调队列解决滑动窗口最值问题的方法。通过维护一个队列来跟踪滑动窗口中的最大值和最小值，该算法可以在O(n)的时间复杂度内求解。文章提供了详细的代码实现，帮助读者理解并掌握这一高效算法。

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Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 54919		Accepted: 15809
Case Time Limit: 5000MS

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.

Window position	Minimum value	Maximum value
[1 3 -1] -3 5 3 6 7	-1	3
1 [3 -1 -3] 5 3 6 7	-3	3
1 3 [-1 -3 5] 3 6 7	-3	5
1 3 -1 [-3 5 3] 6 7	-3	5
1 3 -1 -3 [5 3 6] 7	3	6
1 3 -1 -3 5 [3 6 7]	3	7

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

题目大意：

输出区间内的最值。

思路：

最简单的单调队列的应用。直接看代码好了，简单粗暴。

AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int q[1000005];
int num[1000005];
int xx[1000005];
int n,k;
void the_max()
{
    int dex=-1;int head=0;
    for(int i=0;i<k-1;i++)
    {
        while(dex>=head&&q[dex]<=num[i])
            dex--;
        dex++;
        q[dex]=num[i];
        xx[dex]=i;
    }
    for(int i=k-1;i<n;i++)
    {
        while(dex>=head&&q[dex]<=num[i])
            dex--;
        dex++;
        q[dex]=num[i];
        xx[dex]=i;
        while(xx[head]<=i-k)
            head++;
        cout<<q[head]<<" ";
    }
}
void the_min()
{
    int dex=-1;int head=0;
    for(int i=0;i<k-1;i++)
    {
        while(dex>=head&&q[dex]>=num[i])
            dex--;
        dex++;
        q[dex]=num[i];
        xx[dex]=i;
    }
    for(int i=k-1;i<n;i++)
    {
        while(dex>=head&&q[dex]>=num[i])
            dex--;
        dex++;
        q[dex]=num[i];
        xx[dex]=i;
        while(xx[head]<=i-k)
            head++;
        cout<<q[head]<<" ";
    }
}
int main()
{

    while(~scanf("%d%d",&n,&k))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num[i]);
        }
        the_min();
        cout<<endl;
        the_max();
        cout<<endl;
    }
}