Sliding Window

Time Limit: 12000MS	Memory Limit: 65536K
Total Submissions: 62889	Accepted: 17951
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

題目大意：

求每個視窗的最大最小值。

思路：

區間RMQ問題，這裡溫習（重新學習）一下單調佇列。

單調佇列基礎操作：

①出隊，我們首先將老元素從隊頭依次出隊。

②入隊，我們從後向前掃，找到第一個小於（大於）當前入隊的元素之後，將當前元素放到這個位子後邊，同時刪除所有後邊不必要存在元素。

那麼對應這個題，我們出隊之後入隊然後查詢即可。

Ac程式碼：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
struct node
{
    int val,pos;
} q[3500000];
int a[3500000];
int ans[3500000];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)scanf("%d",&a[i]);
        int s=0,e=-1;
        for(int i=1; i<=n; i++)
        {
            while(s<=e&&q[s].pos<i-m+1)s++;
            while(s<=e&&q[e].val>a[i])e--;
            e++,q[e].pos=i,q[e].val=a[i];
            if(i>=m)ans[i]=q[s].val;
        }
        for(int i=m; i<=n; i++)
        {
            if(i>m)printf(" ");
            printf("%d",ans[i]);
        }
        printf("\n");
        s=0,e=-1;
        for(int i=1; i<=n; i++)
        {
            while(s<=e&&q[s].pos<i-m+1)s++;
            while(s<=e&&q[e].val<a[i])e--;
            e++,q[e].pos=i,q[e].val=a[i];
            if(i>=m)ans[i]=q[s].val;
        }
        for(int i=m; i<=n; i++)
        {
            if(i>m)printf(" ");
            printf("%d",ans[i]);
        }
        printf("\n");
    }
    return 0;
}