POJ 题目2823 Sliding Window（RMQ，固定区间长度）

最新推荐文章于 2019-08-13 20:25:55 发布

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最新推荐文章于 2019-08-13 20:25:55 发布

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原文链接：http://www.cnblogs.com/llguanli/p/8312929.html

本文介绍了一种用于数组处理的滑动窗口算法，该算法能在数组上移动固定大小的窗口并高效地找出每个位置的最大值和最小值。适用于长度不超过10^6的数据处理。

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 46507		Accepted: 13442
Case Time Limit: 5000MS

Description

An array of size n ≤ 10 ⁶ 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

Source

POJ Monthly--2006.04.28, Ikki

固定区间长度一维的数组就好，二维超内存

ac代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
int maxv[1000010],minv[1000010],a[1000010],n,m,k;
void init()
{
	int i,j;
	for(i=1;i<=n;i++)
	{
		minv[i]=a[i];
		maxv[i]=a[i];
	}
	for(i=1;i<=k;i++)
	{
		for(j=1;j+(1<<i)-1<=n;j++)
		{
			minv[j]=min(minv[j],minv[j+(1<<(i-1))]);
			maxv[j]=max(maxv[j],maxv[j+(1<<(i-1))]);
		}
	}
}
int q_max(int l,int r)
{
//	int k=(int)(log((double)(r-l+1))/log(2.0));
	return max(maxv[l],maxv[r-(1<<k)+1]);
}
int q_min(int l,int r)
{
//	int k=(int)(log((double)(r-l+1))/log(2.0));
	return min(minv[l],minv[r-(1<<k)+1]);
}
int main()
{
//	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		k=(int)(log(double(m))/log(2.0));
		init();
		if(m<=n)
		{
			printf("%d",q_min(1,m+1-1));
		}
		for(i=2;i+m-1<=n;i++)
		{
			printf(" %d",q_min(i,m+i-1));
		}
		printf("\n");
		if(m<=n)
		{
			printf("%d",q_max(1,m+1-1));
		}
		for(i=2;i+m-1<=n;i++)
		{
			printf(" %d",q_max(i,m+i-1));
		}
		printf("\n");
	}
}

转载于:https://www.cnblogs.com/llguanli/p/8312929.html