PKU2823 单调队列

滑窗算法

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本文介绍了一种基于滑动窗口的算法，用于确定数组中特定窗口大小的最大值和最小值。通过使用单调队列来高效地解决该问题，并提供了一个C语言实现的例子。

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 14169		Accepted: 3953
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

单调队列的入门练习题 .

关于单调队列的资料 : http://flyfox141.blog.hexun.com/15302431_d.html

代码如下 :

#include<stdio.h> #include<string.h> #define MAXN 1000005 struct Q { int v; int num; }que[MAXN]; int a[MAXN],head,tail; void inmin(int i) { while (head<=tail && que[tail].v>a[i]) tail--; que[++tail].v=a[i]; que[tail].num=i; } void inmax(int i) { while (head<=tail && que[tail].v<a[i]) tail--; que[++tail].v=a[i]; que[tail].num=i; } int main() { int n,k,i; while (scanf("%d%d",&n,&k)!=EOF) { if (n<k) k=n; for (i=1;i<=n;++i) scanf("%d",&a[i]); head=tail=0; memset(que,0,sizeof(que)); for (i=1;i<k;++i) inmin(i); for (i=k;i<=n;++i) { inmin(i); while (head<=tail && que[head].num<i-k+1) head++; printf("%d",que[head].v); if (i!=n) printf(" "); } printf("/n"); head=tail=0; for (i=1;i<k;++i) inmax(i); for (i=k;i<=n;++i) { inmax(i); while (head<=tail && que[head].num<i-k+1) head++; printf("%d",que[head].v); if (i!=n) printf(" "); } printf("/n"); } return 0; }