UESTC_Sliding Window 2015 UESTC Training for Data Structures<Problem K>

K - Sliding Window

Time Limit: 18000/6000MS (Java/Others)     Memory Limit: 131072/131072KB (Java/Others)

Submit Status

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

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 and output

Sample Input	Sample Output
8 3 1 3 -1 -3 5 3 6 7	-1 -3 -3 -3 3 3 3 3 5 5 6 7

Hint

The data used in this problem is unofficial data prepared by love8909. So any mistake here does not imply mistake in the offcial judge data.

解题报告

滑动窗口问题，我们可以在O（1）的时间内得到某个点的答案，就是维护一个单调队列，首先考虑最大值问题，我们考虑 i < j,且a[i] < a[j],显然可以得到a[i]是根本无用的（因为从左往右滑，a[j]未出之前a[i]根本不可能最优),因此我们只需维护一个单调递减的队列的即可,即可在O(1)的时间内得到某个点最优值.

插入时也要维护单调性，不再累述.

最小值同理.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;
const int maxn = 1e6 + 50;
int n,k,q[maxn],h[maxn];

int main(int argc,char *argv[])
{
  scanf("%d%d",&n,&k);
  int front = 0 , rear = 0 ;
  for(int i = 0 ; i < n ; ++ i)
   scanf("%d",&h[i]);
  // Judge
  if (k >= n)
   k = n;
  // Init
  q[rear++] = 0;
  for(int i = 1 ; i < k ; ++ i)
   {
          while(h[i] <= h[q[rear-1]] && front < rear)
         rear--;
       q[rear++] = i; 
   }
  printf("%d",h[q[front]]);
  for(int i = k ; i < n ; ++ i)
   {
          while(front < rear && i - q[front] >= k)
           front++;
          while(h[i] <= h[q[rear-1]] && front < rear)
         rear--;
       q[rear++] = i;
       printf(" %d",h[q[front]]);
   }
  printf("
");
  // ReInit
  front = 0 , rear = 0 , q[rear++] = 0;
  for(int i = 1 ; i < k ; ++ i)
   {
          while(h[i] >= h[q[rear-1]] && front < rear)
         rear--;
       q[rear++] = i; 
   }
  printf("%d",h[q[front]]);
  for(int i = k ; i < n ; ++ i)
   {
          while(front < rear && i - q[front] >= k)
           front++;
          while(h[i] >= h[q[rear-1]] && front < rear)
         rear--;
       q[rear++] = i;
       printf(" %d",h[q[front]]);
   }
  printf("
");
  return 0;
}