POJ - 2823 Sliding Window（单调队列）

POJ - 2823 Sliding Window（单调队列）

滑动窗口的最值问题

http://poj.org/problem?id=2823

 

链接：https://ac.nowcoder.com/acm/problem/51001
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

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. 

输入描述:

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.

输出描述:

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.

示例1

输入

8 3
1 3 -1 -3 5 3 6 7

输出

-1 -3 -3 -3 3 3
3 3 5 5 6 7

 

 

 

 

 1 #include <bits/stdc++.h>
 2 typedef long long LL;
 3 #define pb push_back
 4 #define mst(a) memset(a,0,sizeof(a))
 5 const int INF = 0x3f3f3f3f;
 6 const double eps = 1e-8;
 7 const int mod = 1e9+7;
 8 const int maxn = 1e6+10;
 9 using namespace std;
10 
11 int a[maxn];
12 vector<int> ans1,ans2;
13 deque<int> qe1,qe2;
14 
15 int main()
16 {
17     #ifdef DEBUG
18     freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout);
19     #endif
20     
21     int n,m;
22     scanf("%d %d",&n,&m);
23     for(int i=1;i<=n;i++)
24         scanf("%d",&a[i]);
25     for(int i=1;i<=n;i++)
26     {
27         //维护最小值队列qe1
28         if(!qe1.empty()&&qe1.front()<=i-m) qe1.pop_front();
29         if(qe1.empty()) qe1.push_back(i);
30         else
31         {
32             if(a[i]<=a[qe1.front()])
33             {
34                 qe1.clear();
35                 qe1.push_back(i);
36             }
37             else
38             {
39                 if(qe1.size()==1) qe1.push_back(i);
40                 else
41                 {
42                     if(a[i]>=a[qe1.back()]) qe1.push_back(i);
43                     else
44                     {
45                         int t = qe1.back();
46                         while(a[i]<a[t])
47                         {
48                             qe1.pop_back();
49                             t = qe1.back();
50                         }
51                         qe1.push_back(i);
52                     }
53                 }
54             }
55         }
56         //维护最大值队列qe2
57         if(!qe2.empty()&&qe2.front()<=i-m) qe2.pop_front();
58         if(qe2.empty()) qe2.push_back(i);
59         else
60         {
61             if(a[i]>=a[qe2.front()])
62             {
63                 qe2.clear();
64                 qe2.push_back(i);
65             }
66             else
67             {
68                 if(qe2.size()==1) qe2.push_back(i);
69                 else
70                 {
71                     if(a[i]<=a[qe2.back()]) qe2.push_back(i);
72                     else
73                     {
74                         int t = qe2.back();
75                         while(a[i]>a[t])
76                         {
77                             qe2.pop_back();
78                             t = qe2.back();
79                         }
80                         qe2.push_back(i);
81                     }
82                 }
83             }
84         }
85         
86         if(i>=m)
87         {
88             ans1.push_back(a[qe1.front()]);
89             ans2.push_back(a[qe2.front()]);
90         }
91     }
92 
93     for(int i=0;i<=n-m;i++)
94         printf(i==n-m?"%d\n":"%d ",ans1[i]);
95     for(int i=0;i<=n-m;i++)
96         printf(i==n-m?"%d\n":"%d ",ans2[i]);
97     
98     return 0;
99 }

 

 

 

 

 

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posted @ 2020-07-01 14:43  jiamian22  阅读(152)  评论(0)    收藏  举报