POJ 2823 Sliding Window（单调队列入门题）

 

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 67218		Accepted: 19088
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

Source

POJ Monthly--2006.04.28, Ikki

这个问题用RMQ或者倍增法能在o（nlogn）内解决。 但是用双端队列（单调队列）能在o（n）内解决。 求最小值：建立一个单调递增队列，元素从左到右依次入队，入队之前必须从队列发问开始删除那些比当前入队元素大或者相等的元素，直到遇到一个比当前入队元素小的元素，或者队列为空为止。若此时队列的大小超过窗口值，则从队头删除元素，直到队列大小小入窗口值为止。然后把当前元素插入队尾。

每滑动一次，取队列的队首元素作为这一区间范围的最小值。并把超过区间范围的最小值删除。若有更小的值入队，则把队列里其它比它大的删了，因为这个最小值在它们后面，又比它们小，所以它们没有存在的意义了。

 

#include<stdio.h>
#include<string.h>
#include<stdio.h>  
#include<string.h>  
#include<stdlib.h>  
#include<queue>  
#include<stack>  
#include<math.h>  
#include<vector>  
#include<map>  
#include<set>  
#include<cmath>  
#include<string>  
#include<algorithm>  
#include<iostream>  
#include<string.h>
#include<algorithm>
#include<vector>
#include<stdio.h>
#include<cstdio>
#include<time.h>
#include<stack>
#include<queue>
#include<deque>
#include<map>
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
int mi[1000005];
int ma[1000005];
int a[1000005];
deque<int>q;
int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    while(!q.empty ()) q.pop_back();
    for(int i=1;i<=n;i++)
    {
        while(!q.empty ()&&a[q.back ()]>=a[i]) q.pop_back();
        if(q.empty ()) mi[i]=i;
        else mi[i]=q.front ();
        q.push_back (i);
        if(q.front ()==i-k+1) q.pop_front ();
    }

    while(!q.empty ()) q.pop_back();
    for(int i=1;i<=n;i++)
    {
        while(!q.empty ()&&a[q.back ()]<=a[i]) q.pop_back();
        if(q.empty ()) ma[i]=i;
        else ma[i]=q.front ();
        q.push_back (i);
        if(q.front ()==i-k+1) q.pop_front ();
    }

    for(int i=k;i<=n;i++)
    {
        printf("%d",a[mi[i]]);
        if(i!=n) printf(" ");
    }
    printf("
");
    for(int i=k;i<=n;i++)
    {
        printf("%d",a[ma[i]]);
        if(i!=n) printf(" ");
    }
    return 0;
}