POJ 2823 Sliding Window （滑动窗口的最值问题 ）

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.

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;
}