题解报告：poj 2823 Sliding Window（单调队列）

Posted 2021-01-01 acgoto

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
解题思路：这题不止一种做法，初学单调队列，这题作为入门题再适合不过了。单调队列维护固定区间长度的最值，求区间长为k中的最大值用单调递减队列，求区间长为k中的最小值用单调递增队列。单调队列和单调栈十分相似，但又有区别。相关讲解：单调队列总结。时间复杂度是O(n)。还有一点这道题要用C++提交，用G++会超时，原因不清楚=_=||。
AC代码(6829ms)：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string.h>
 5 #include<deque>
 6 using namespace std;
 7 const int maxn=1e6+5;
 8 int n,k,a[maxn],minv[maxn],maxv[maxn];
 9 deque<int> dq1,dq2;
10 int main(){
11     while(~scanf("%d%d",&n,&k)){
12         for(int i=0;i<n;++i)scanf("%d",&a[i]);
13         dq1.clear();dq2.clear();memset(minv,0,sizeof(minv));memset(maxv,0,sizeof(maxv));
14         for(int i=0;i<n;++i){
15             while(!dq1.empty()&&a[dq1.back()]>=a[i])dq1.pop_back();//维护窗口最小值，单调递增队列
16             while(!dq2.empty()&&a[dq2.back()]<=a[i])dq2.pop_back();//维护窗口最大值，单调递减队列
17             dq1.push_back(i);dq2.push_back(i);
18             if(i-k+1>=0){//至少从第k个数开始才有区间最值
19                 minv[i-k+1]=a[dq1.front()],maxv[i-k+1]=a[dq2.front()];//直接取队首在当前窗口的最值
20                 if(dq1.front()==i-k+1)dq1.pop_front();//如果队首下标已经是最大区间(一共k个元素)的左端点值，则将其弹出，窗口向右移动
21                 if(dq2.front()==i-k+1)dq2.pop_front();
22             }
23         }
24         for(int i=0;i<=n-k;++i)//输出n-k+1个窗口中的最值
25             printf("%d%c",minv[i],i==n-k?‘
‘:‘ ‘);
26         for(int i=0;i<=n-k;++i)
27             printf("%d%c",maxv[i],i==n-k?‘
‘:‘ ‘);
28     }
29     return 0;
30 }