zoukankan      html  css  js  c++  java
  • POJ 2823 Sliding Window

    POJ 2823 Sliding Window

    Description

    An array of size 6is given to you. There is a sliding window of sizekwhich is moving from the very left of the array to the very right. You can only see theknumbers 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]

    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
    

    题解:

    滑动窗口。

    代码:

    #include<cstdio>
    #include<deque>
    using namespace std;
    const int maxn=1e6+6;
    int n,k;
    int a[maxn];
    deque<pair<int,int> > q;
    int main()
    {
    	scanf("%d%d",&n,&k);
    	for(int i=1;i<=n;i++)
    		scanf("%d",&a[i]);
    	for(int i=1;i<=n;i++)
    	{
    		while(!q.empty() && a[i]<q.back().second)
    			q.pop_back();
    		q.push_back(make_pair(i,a[i]));
    		while(q.front().first<=i-k)
    			q.pop_front();
    		if(i>=k)
    			printf("%d ",q.front().second);
    	}
    	puts("");
    	q.clear();
    	for(int i=1;i<=n;i++)
    	{
    		while(!q.empty() && a[i]>q.back().second)
    			q.pop_back();
    		q.push_back(make_pair(i,a[i]));
    		while(q.front().first<=i-k)
    			q.pop_front();
    		if(i>=k)
    			printf("%d ",q.front().second);
    	}
    	puts("");
    	return 0;
    }
    
  • 相关阅读:
    POJ 2739:Sum of Consecutive Prime Numbers(Two pointers)
    POJ 2566:Bound Found(Two pointers)
    Codeforces 528D Fuzzy Search(FFT)
    挑战程序设计竞赛 3.1 不光是查找值!“二分搜索”
    POJ 3484 Showstopper(二分答案)
    POJ 1759 Garland(二分答案)
    POJ 3662 Telephone Lines(二分答案+SPFA)
    POJ 3579 Median(二分答案+Two pointers)
    POJ 3111 K Best(二分答案)
    POJ 2976 Dropping tests(二分答案)
  • 原文地址:https://www.cnblogs.com/fusiwei/p/13926032.html
Copyright © 2011-2022 走看看