Sliding Window
| Time Limit: 12000MS | Memory Limit: 65536K | |
| Total Submissions: 41264 | Accepted: 12229 | |
| 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.
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
我按照自己的思路进行了部分优化模拟，仍然超时。
代码：
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <deque>
using namespace std;
struct node
{
int mi;
int ma;
}q[1000004];
int a[1000004];
int main()
{
int n, k;
int i, j;
scanf("%d %d", &n, &k);
for(int i=0; i<n; i++)
scanf("%d", &a[i] );
int dd=a[0], ff=a[0];
for(i=1; i<k; i++)
{
if(a[i]>dd) dd=a[i]; //max
if(a[i]<ff) ff=a[i]; //min
}
int e=0;
q[e].ma=dd; q[e++].mi=ff;
int pos;
for(j=k; j<n; j++ )// 遍历这些序列
{
if(a[j]>dd) dd=a[j];
if(a[j]<ff) ff=a[j]; //判断是不是要更新
pos=j-k;
if(a[pos]>dd && a[pos]<ff )
continue;
else
{
int p, w;
if(a[pos]==dd) //起点是==最大值 更新最大值
{
p=a[pos+1];
for(i=pos+2; i<=j; i++)
{
p=max(p, a[i]);
}
dd=p;
}
else if(a[pos]==ff) //起点是==最小值 更新最小值
{
w=a[pos+1];
for(i=pos+2; i<=j; i++)
{
w=min(w, a[i]);
}
ff=w;
}
q[e].ma=dd; q[e++].mi=ff;
}
}
for(i=0; i<e; i++)
{
if(i==e-1)
printf("%d\n", q[i].mi);
else
printf("%d ", q[i].mi);
}
for(i=0; i<e; i++)
{
if(i==e-1)
printf("%d\n", q[i].ma );
else
printf("%d ", q[i].ma );
}
return 0;
}