poj-2823 Sliding Window

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Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 79083		Accepted: 22340
Case Time Limit: 5000MS

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

Source

POJ Monthly--2006.04.28, Ikki

..这题一定要用c++ 否则用G++ TLE

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <stdio.h>
 5 #include <cstring>
 6 #include <string>
 7 #include <cstdlib>
 8 #include <queue>
 9 #include <stack>
10 #include <set>
11 #include <vector>
12 #include <map>
13 #include <list>
14 #include <iomanip>
15  #include <fstream>
16 using namespace std;
17 typedef long long ll;
18 const int maxn=1e6;
19 int a[maxn+1],q1[maxn+1],q2[maxn+1],maxnn[maxn+1],minn[maxn+1]; 
20 int main()
21 {
22     int n,k,x;
23     scanf("%d%d",&n,&k);
24     for(int i=1;i<=n;++i)
25     {
26         scanf("%d",&a[i]);
27     }
28     int h1=1,h2=1,t1=0,t2=0;
29     for(int i=1;i<=n;++i)
30     {    
31         while(h1<=t1&&a[q1[t1]]<=a[i])//当长度为k的队列头部没有超过尾部的时候，并且这个队列尾部的下标所代表的值比当前遍历到的a[i]要小，那么队尾收缩 
32             t1--;
33         q1[++t1]=i;//队列尾指针向后移动一位，并且队列尾部下标为当前遍历到的下标i。    
34         while(i-k>=q1[h1])//当前的 i- 滑块长度k  所得到的下标 不小于当前头指针下标。 5-3 =2  
35             h1++;       //那么长度为k的队列头指针向后移动，直到  h~i 的长度不超过要求的滑块长度k。 删队首 
36         maxnn[i]=a[q1[h1]]; //当前滑块的最大值就是长度为k的队列头指针所指向的下标所代表的值。
37         
38         
39         while(h2<=t2&&a[q2[t2]]>=a[i])
40             t2--;
41         q2[++t2]=i;
42         while(i-k>=q2[h2])
43             h2++;
44         minn[i]=a[q2[h2]];
45         
46     }
47     
48     for(int i=k;i<=n;++i)
49         printf("%d ",minn[i]);
50     printf("\n");
51     for(int i=k;i<=n;++i)        
52         printf("%d ",maxnn[i]);
53 
54     return 0;
55 }

View Code

 

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转载于:https://www.cnblogs.com/greenaway07/p/11217565.html