POJ 题目2823 Sliding Window（RMQ，固定区间长度）

Sliding Window

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

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("
");
		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("
");
	}
}