Sliding Window 单调队列求区间内的最值

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

***************************************************************************************************************************

单调队列，求区间内的最值（注意是多个定长区间）

***************************************************************************************************************************

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstdio>
 6 #include<algorithm>
 7 using namespace std;
 8 int a[1000001];
 9 int que[100000001];
10 int n,k;
11 void Min()//单调递增求最小值
12 {
13     int head=1,tail=0;
14     int it;
15     for(it=0;it<k-1;it++)
16     {
17         while(head<=tail&&a[que[tail]]>=a[it])tail--;
18         tail++;
19         que[tail]=it;
20     }
21     for(it=k-1;it<n;it++)
22     {
23         while(head<=tail&&a[que[tail]]>=a[it])tail--;
24         tail++;
25         que[tail]=it;
26         while(que[head]<it-k+1)head++;
27         printf("%d",a[que[head]]);
28         printf("%c",((it==n-1)?'\n':' '));
29 
30     }
31 }
32 void Max()//单调递减，求最大值
33 {
34     int head=1,tail=0;
35     int it;
36     for(it=0;it<k-1;it++)
37     {
38         while(head<=tail&&a[que[tail]]<=a[it])tail--;
39         tail++;
40         que[tail]=it;
41     }
42     for(it=k-1;it<n;it++)
43     {
44         while(head<=tail&&a[que[tail]]<=a[it])tail--;
45         tail++;
46         que[tail]=it;
47         while(que[head]<it-k+1)head++;
48         printf("%d",a[que[head]]);
49         printf("%c",((it==n-1)?'\n':' '));
50 
51     }
52 }
53 int main()
54 {
55     int i,j;
56     while(scanf("%d%d",&n,&k)!=EOF)
57     {
58         for(i=0;i<n;i++)
59         scanf("%d",&a[i]);
60         Min();
61         Max();
62     }
63     return 0;
64 
65 }

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转载于:https://www.cnblogs.com/sdau--codeants/p/3450728.html