POJ 2823 Sliding Window（单调队列）

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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。

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 61914		Accepted: 17733
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：

滑动窗口，一次只能看k个数字，求每次窗口中的最大值最小值。

POINT：

单调队列。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <algorithm>
using namespace std;
#define  LL long long
const int N = 1000000+10;
struct node
{
    int x,y;
}v[N];
int a[N],n,k;
int ans[N];
void getmax()
{
    int cnt=0;
    int head=1,end=0;
    for(int i=1;i<k;i++)
    {
        while(head<=end&&v[end].x<=a[i]) end--;
        v[++end].x=a[i],v[end].y=i;
    }
    for(int i=k;i<=n;i++)
    {
        while(head<=end&&v[end].x<=a[i]) end--;
        v[++end].x=a[i],v[end].y=i;
        while(i-v[head].y>=k) head++;
        ans[++cnt]=v[head].x;
    }
    for(int i=1;i<=cnt;i++)
    {
        if(i-1) printf(" ");
        printf("%d",ans[i]);
    }
    printf("\n");
}
void getmin()
{
    int cnt=0;
    int head=1,end=0;
    for(int i=1;i<k;i++)
    {
        while(head<=end&&v[end].x>=a[i]) end--;
        v[++end].x=a[i],v[end].y=i;
    }
    for(int i=k;i<=n;i++)
    {
        while(head<=end&&v[end].x>=a[i]) end--;
        v[++end].x=a[i],v[end].y=i;
        while(i-v[head].y>=k) head++;
        ans[++cnt]=v[head].x;
    }
    for(int i=1;i<=cnt;i++)
    {
        if(i-1) printf(" ");
        printf("%d",ans[i]);
    }
    printf("\n");
    
}
int main()
{
    scanf("%d %d",&n,&k);
    
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        getmin();
        getmax();
    
}