正文
POJ 题目2823 Sliding Window（RMQ，固定区间长度）
小程序：扫一扫查出行【扫一扫了解最新限行尾号】
复制小程序
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.
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");
}
}
