POJ 2838 单调队列

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 55309		Accepted: 15911
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

题意：

N个数一串，一个长度为K的窗口，窗口开始在最左边，每次向右移动一格，问每次窗口内最大值和最小值。

代码：

基本单调队列

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 int n,k;
 int a[];
 int ddq[];
 void min()
 {
     int head=,tail=;
     for(int i=;i<k-;i++)
     {
         while(head<=tail&&a[ddq[tail]]>=a[i])
         tail--;
         tail++;
         ddq[tail]=i;
     }
     for(int i=k-;i<n;i++)
     {
         while(head<=tail&&a[ddq[tail]]>=a[i])
         tail--;
         tail++;
         ddq[tail]=i;
         while(ddq[head]<i-k+)
         head++;
         printf("%d",a[ddq[head]]);
         if(i==n-) printf("\n");
         else printf(" ");

     }
 }
 void max()
 {
     int head=,tail=;
     for(int i=;i<k-;i++)
     {
         while(head<=tail&&a[ddq[tail]]<=a[i])
         tail--;
         tail++;
         ddq[tail]=i;
     }
     for(int i=k-;i<n;i++)
     {
         while(head<=tail&&a[ddq[tail]]<=a[i])
         tail--;
         tail++;
         ddq[tail]=i;
         while(ddq[head]<i-k+)
         head++;
         printf("%d",a[ddq[head]]);
         if(i==n-)
         printf("\n");
         else printf(" ");
     }
 }
 int main()
 {
     while(scanf("%d%d",&n,&k)!=EOF)
     {
         memset(ddq,,sizeof(ddq));
         memset(a,,sizeof(a));
         for(int i=;i<n;i++)
         scanf("%d",&a[i]);
         min();
         max();
     }
     return ;
 }