POJ 2823 Sliding Window

Sliding Window

 

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

题意：
　　一个长度为n的序列，用长度为看k的窗口在上面移动，求窗口中数字最大为多少。
分析：
　　单调队列求，入门题目收下了。、

# include <iostream>
using namespace std;

const int MAX = 1000010;
int a[MAX];
int q[MAX];
int p[MAX];
int Min[MAX];
int Max[MAX];

int n, k;

void get_Min()
{
    int head = 1, tail = 0;
    for(int i = 0; i < k - 1; i++)
    {
        while(head <= tail && q[tail] >= a[i])
            tail--;
        tail++;
        q[tail] = a[i];
        p[tail] = i;
    }
    for(int i = k - 1; i < n; i++)
    {
        while(head <= tail && q[tail] >= a[i])
            tail--;
        tail++;
        q[tail] = a[i];
        p[tail] = i;
        while(p[head] < i - k + 1)
        {
            head++;
        }
        Min[i - k + 1] = q[head];
    }
}
void get_Max()
{
    int head = 1, tail = 0;
    for(int i = 0; i < k - 1; i++)
    {
        while(head <= tail && q[tail] <= a[i])
            tail--;
        tail++;
        q[tail] = a[i];
        p[tail] = i;
    }
    for(int i = k - 1; i < n; i++)
    {
        while(head <= tail && q[tail] <= a[i])
            tail--;
        tail++;
        q[tail] = a[i];
        p[tail] = i;
        while(p[head] < i - k + 1)
        {
            head++;
        }
        Max[i - k + 1] = q[head];
    }
}



int main()
{
    scanf("%d%d", &n, &k);
    for(int i = 0; i < n; i++)
        scanf("%d", &a[i]);
    get_Min();
    get_Max();
    for(int i = 0; i < n - k + 1; i++)
    {
        if(i == 0)
            printf("%d", Min[i]);
        else
            printf(" %d", Min[i]);
    }
    printf("
");
    for(int i = 0; i < n - k + 1; i++)
    {
        if(i == 0)
            printf("%d", Max[i]);
        else
            printf(" %d", Max[i]);
    }
    printf("
");
    return 0;
}