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

這個問題用RMQ或者倍增法能在o（nlogn）內解決。 但是用雙端佇列（單調佇列）能在o（n）內解決。 求最小值：建立一個單調遞增佇列，元素從左到右依次入隊，入隊之前必須從佇列發問開始刪除那些比當前入隊元素大或者相等的元素，直到遇到一個比當前入隊元素小的元素，或者佇列為空為止。若此時佇列的大小超過視窗值，則從隊頭刪除元素，直到佇列大小小入視窗值為止。然後把當前元素插入隊尾。

每滑動一次，取佇列的隊首元素作為這一區間範圍的最小值。並把超過區間範圍的最小值刪除。若有更小的值入隊，則把佇列裡其它比它大的刪了，因為這個最小值在它們後面，又比它們小，所以它們沒有存在的意義了。

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