单调队列——Poj Sliding Window
来源:互联网 发布:云计算面临的安全问题 编辑:程序博客网 时间:2024/06/05 09:28
Sliding Window
Time Limit: 12000MS Memory Limit: 65536KTotal Submissions: 33379 Accepted: 9924Case 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 knumbers 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 -13 1 [3 -1 -3] 5 3 6 7 -33 1 3 [-1 -3 5] 3 6 7 -35 1 3 -1 [-3 5 3] 6 7 -35 1 3 -1 -3 [5 3 6] 7 36 1 3 -1 -3 5 [3 6 7]37
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
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 31 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 33 3 5 5 6 7
题意:
给定含有n个元素的无序序列a[],和一个整数k,要求求出a[]中,从左向右每连续k个元素组成的序列中的最小值(或最大值),这样的值可能有1个或n-k+1个。
思路:
利用单调队列来求解,需要注意的是,单调队列的最大长度为k。单调队列
代码:
#include <stdio.h>#define MAXN 1000000int mq[MAXN+5];//单调队列,存储元素的索引int f;//队首指针int r;//队尾指针int a[MAXN+5];//数字串int n;//数字个数int k;//滑动窗口大小void Push_Asc(int i)//按升序进队{while (r > f && a[i] < a[mq[r-1]]){r--;}mq[r++] = i;}void Push_Desc(int i)//按降序进队{while (r > f && a[i] > a[mq[r-1]]){r--;}mq[r++] = i;}inline int Front(void){return mq[f];}inline bool IsEmpty(void){return f == r;}inline void Pop(void){f++;}void SlidingWindow(bool ascending){f = r = 0;//初始化队列void (*Push)(int) = ascending ? Push_Asc : Push_Desc;//判断单调队列是上升还是下降int i;for (i = 1; i <= k && i <= n; i++)//让前k个数进队,k有可能大于n{Push(i);}printf("%d", a[Front()]);for (; i <= n; i++){while (!IsEmpty() && Front() + k <= i)//弹出不在滑动窗口内的元素{Pop();}Push(i);printf(" %d", a[Front()]);}putchar('\n');}int main(void){while (scanf("%d%d", &n, &k) != EOF){int i;for (i = 1; i <= n; i++){scanf("%d", &a[i]);}SlidingWindow(true);SlidingWindow(false);}return 0;}
- 单调队列——Poj Sliding Window
- 单调队列+IO优化 POJ——Sliding Window
- POJ2823.Sliding Window——单调队列
- poj 2823 Sliding Window (单调队列)
- POJ 2823 Sliding Window 堆 / 单调队列
- POJ 2823 Sliding Window 单调队列
- poj 2823 Sliding Window 【单调队列】
- poj 2823 Sliding Window(单调队列)
- POJ 2823 Sliding Window(单调队列)
- poj 2823 Sliding Window(简单单调队列)
- POJ 2823 Sliding Window(单调队列)
- POJ 2823 Sliding Window(单调队列)
- poj 2823 Sliding Window(单调队列)
- POJ 2823 Sliding Window(单调队列)
- poj 2823 Sliding Window(单调队列)
- POj 2823 Sliding Window 单调队列
- poj 2823 Sliding Window ( 单调队列 )
- 单调队列--poj-2823-Sliding Window
- 树的判定(并查集)
- java打印机的一个问题的解决办法
- 栈:顺序栈和链式栈
- cf 359C
- css实现div自动添加纵向滚动条
- 单调队列——Poj Sliding Window
- 黑马程序员---回顾之java多线程技术
- 数据结构与算法实验题 6.1 鼹鼠掘土挑战赛
- 内存屏障 - MemoryBarrier
- proxy-target-class="true" 与proxy-target-class="false"的区别
- java导出excel报错 执行InputStream is = new FileInputStream()
- opencv学习_6 (灰度直方图)
- ForSecondWork-No.2:C/C++笔试题回忆并整理
- tree(nyoj)