POJ 2823 Sliding Window

Sliding Window

 

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.

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

題意：
　　一個長度爲n的序列，用長度爲看k的窗口在上面移動，求窗口中數字最大爲多少。
分析：
　　單調隊列求，入門題目收下了。、

 1 # include <iostream>
 2 using namespace std;
 3 
 4 const int MAX = 1000010;
 5 int a[MAX];
 6 int q[MAX];
 7 int p[MAX];
 8 int Min[MAX];
 9 int Max[MAX];
10 
11 int n, k;
12 
13 void get_Min()
14 {
15     int head = 1, tail = 0;
16     for(int i = 0; i < k - 1; i++)
17     {
18         while(head <= tail && q[tail] >= a[i])
19             tail--;
20         tail++;
21         q[tail] = a[i];
22         p[tail] = i;
23     }
24     for(int i = k - 1; i < n; i++)
25     {
26         while(head <= tail && q[tail] >= a[i])
27             tail--;
28         tail++;
29         q[tail] = a[i];
30         p[tail] = i;
31         while(p[head] < i - k + 1)
32         {
33             head++;
34         }
35         Min[i - k + 1] = q[head];
36     }
37 }
38 void get_Max()
39 {
40     int head = 1, tail = 0;
41     for(int i = 0; i < k - 1; i++)
42     {
43         while(head <= tail && q[tail] <= a[i])
44             tail--;
45         tail++;
46         q[tail] = a[i];
47         p[tail] = i;
48     }
49     for(int i = k - 1; i < n; i++)
50     {
51         while(head <= tail && q[tail] <= a[i])
52             tail--;
53         tail++;
54         q[tail] = a[i];
55         p[tail] = i;
56         while(p[head] < i - k + 1)
57         {
58             head++;
59         }
60         Max[i - k + 1] = q[head];
61     }
62 }
63 
64 
65 
66 int main()
67 {
68     scanf("%d%d", &n, &k);
69     for(int i = 0; i < n; i++)
70         scanf("%d", &a[i]);
71     get_Min();
72     get_Max();
73     for(int i = 0; i < n - k + 1; i++)
74     {
75         if(i == 0)
76             printf("%d", Min[i]);
77         else
78             printf(" %d", Min[i]);
79     }
80     printf("\n");
81     for(int i = 0; i < n - k + 1; i++)
82     {
83         if(i == 0)
84             printf("%d", Max[i]);
85         else
86             printf(" %d", Max[i]);
87     }
88     printf("\n");
89     return 0;
90 }

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