POJ 2823 Sliding Window

Sliding Window

Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 30061 Accepted: 8944Case 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 positionMinimum valueMaximum value[1  3  -1] -3  5  3  6  7 -13 1 [3  -1  -3] 5  3  6  7 -33 1  3 [-1  -3  5] 3  6  7 -35 1  3  -1 [-3  5  3] 6  7 -35 1  3  -1  -3 [5  3  6] 7 36 1  3  -1  -3  5 [3  6  7]37

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 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 

题意：

就是求长度为k的窗口里面的最大值和最小值

解析：

假如说是直接枚举的话，会超时。所以说用单调队列来求，关于单调队列就是队列里面的所有的都是单调递增或递减的，所以说正好适合这样的题目。

#include<stdio.h>#include<string.h>int num[1001000],que[1001000];int n,k;void fxlow()//求最小值{int i,head = 0,rear = 0;for(i = 1; i <= n ; i ++){while(rear >= head && num[i] < num[que[rear]])//凡是比num[i]大的队尾全部舍去rear--;que[++rear] = i;//只保留编号，这样的话下面就可以直接判断了if(que[head] < i - k + 1)//这就是用来判断队列长度是否超过khead++;if(i == k)printf("%d",num[que[head]]);else if(i > k)printf(" %d",num[que[head]]);}printf("\n");}void fxhigh()//求最大值{int i,head = 0,rear = 0;for(i = 1; i <= n ; i ++){while(rear >= head && num[i] > num[que[rear]])rear--;que[++rear] = i;//只保留编号if(que[head] < i - k + 1)//这就是用来判断队列长度是否超过khead++;if(i == k)printf("%d",num[que[head]]);else if(i > k)printf(" %d",num[que[head]]);}printf("\n");} int main() {while(scanf("%d%d",&n,&k) != EOF){memset(que,0,sizeof(que));memset(num,0,sizeof(num));int i;for(i = 1 ; i <= n ;i ++)scanf("%d",&num[i]);fxlow();fxhigh();} }