Sliding Window
Description
An array of size
N≤106 is given to you. There is a sliding window of size
kWhich is moving from the very left of the array to the very right. can only see the
kNumbers in the window. Each of the sliding window moves rightwards by one position. Following is an example:
The array is[1 3-1-3 5 3 6 7], and
kis 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 the lines. The first line contains integers
Nand
kWhich is the lengths of the array and the sliding window. There is
NIntegers in the second line.
Output
There is lines in the output. The first line gives the minimum values in the windows at each position, from left to right, respectively. The second line gives the maximum values.
Sample Input
8 31 3-1-3 5 3 6 7
Sample Output
-1-3-3-3 3 33 3 5 5 6 7
Test instructions
A sequence of length n, with the length of the window to see K on the top of the move, to find the maximum number of Windows.
Analysis:
The monotonous queue asks, the entry question has received. 、
1# include <iostream>2 using namespacestd;3 4 Const intMAX =1000010;5 intA[max];6 intQ[max];7 intP[max];8 intMin[max];9 intMax[max];Ten One intN, K; A - voidget_min () - { the intHead =1, tail =0; - for(inti =0; I < K-1; i++) - { - while(Head <= tail && Q[tail] >=A[i]) +tail--; -tail++; +Q[tail] =A[i]; AP[tail] =i; at } - for(inti = k-1; I < n; i++) - { - while(Head <= tail && Q[tail] >=A[i]) -tail--; -tail++; inQ[tail] =A[i]; -P[tail] =i; to while(P[head] < I-k +1) + { -head++; the } *Min[i-k +1] =Q[head]; $ }Panax Notoginseng } - voidGet_max () the { + intHead =1, tail =0; A for(inti =0; I < K-1; i++) the { + while(Head <= tail && Q[tail] <=A[i]) -tail--; $tail++; $Q[tail] =A[i]; -P[tail] =i; - } the for(inti = k-1; I < n; i++) - {Wuyi while(Head <= tail && Q[tail] <=A[i]) thetail--; -tail++; WuQ[tail] =A[i]; -P[tail] =i; About while(P[head] < I-k +1) $ { -head++; - } -Max[i-k +1] =Q[head]; A } + } the - $ the intMain () the { thescanf"%d%d", &n, &k); the for(inti =0; I < n; i++) -scanf"%d", &a[i]); in get_min (); the Get_max (); the for(inti =0; I < N-k +1; i++) About { the if(i = =0) theprintf"%d", Min[i]); the Else +printf"%d", Min[i]); - } theprintf"\ n");Bayi for(inti =0; I < N-k +1; i++) the { the if(i = =0) -printf"%d", Max[i]); - Else theprintf"%d", Max[i]); the } theprintf"\ n"); the return 0; -}
View Code
POJ 2823 Sliding Window