poj 2823 Sliding Window(单调队列)

最新推荐文章于 2021-01-25 20:02:52 发布

Timeclimber

最新推荐文章于 2021-01-25 20:02:52 发布

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分类专栏：单调队列文章标签：单调队列 poj

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本文介绍了一种利用滑动窗口解决最大值与最小值问题的高效算法，并通过具体实例展示了如何使用单调队列来实现这一过程。适用于数组大小不超过10^6的情况。

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Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 64320		Accepted: 18327
Case Time Limit: 5000MS

Description

An array of size n ≤ 10 ⁶ 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

Source

POJ Monthly--2006.04.28, Ikki

题意：一个可移动的窗口长度为k，问在窗口从最左端移动到最右端每次移动时，最大值和最小值？

思路：单调队列。(学习了单调队列发现和单调栈思路挺像的，只不过每回取值都是取头所以只能用数组模拟)

这题用G++交会超时，改用C++交就可以A了。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e6+5;
int a[maxn];
int q1[maxn],q2[maxn];
int ans1[maxn],ans2[maxn];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        int head1=1,tail1=1,head2=1,tail2=1,j=1;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            while(head1<tail1&&a[q1[tail1-1]]>a[i]) tail1--;
            q1[tail1++]=i;
            if(q1[head1]<i-k+1) head1++;
            if(i>=k) ans1[j]=a[q1[head1]];

            while(head2<tail2&&a[q2[tail2-1]]<a[i]) tail2--;
            q2[tail2++]=i;
            if(q2[head2]<i-k+1) head2++;
            if(i>=k) ans2[j]=a[q2[head2]];
            if(i>=k) j++;
        }
        for(int i=1;i<j;i++)
            printf("%d ",ans1[i]);
        printf("\n");
        for(int i=1;i<j;i++)
            printf("%d ",ans2[i]);
    }
    return 0;
}