POJ 2823 Sliding Window

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 35486		Accepted: 10492
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

PS:单调队列水体。

妈蛋poj还卡scanf，G++超时，换c++输入会快些

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int MAX = 1e6+10;
 6 int a[MAX],b[MAX],pos1[MAX],pos2[MAX];
 7 int ans1[MAX],ans2[MAX];
 8 int main()
 9 {
10     int x,n,k; int cur1,cur2;
11     scanf("%d %d",&n,&k);
12     int rear=0,front=0; cur1=cur2=0;
13     int rear2=0,front2=0;
14     for(int i=0;i<n;i++)
15     {
16         scanf("%d",&x);
17         while(front<rear&&a[rear-1]>x) rear--;
18         a[rear]=x; pos1[rear++]=i;
19         while(front2<rear2&&b[rear2-1]<x) rear2--;
20         b[rear2]=x; pos2[rear2++]=i;
21         if(i>=k-1)
22         {
23             ans1[cur1++]=a[front];
24             ans2[cur2++]=b[front2];
25             if(i-pos1[front]+1>=k) front++;
26             if(i-pos2[front2]+1>=k) front2++;
27         }
28     }
29     for(int i=0;i<cur1;i++)
30     {
31         if(i) printf(" %d",ans1[i]);
32         else printf("%d",ans1[i]);
33     }
34     printf("
");
35     for(int i=0;i<cur2;i++)
36     {
37         if(i) printf(" %d",ans2[i]);
38         else printf("%d",ans2[i]);
39     }
40     printf("
");
41 
42     return 0;
43 }
mycode