Description For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far. Input The first line of input contains…

题目链接:http://poj.org/problem?id=3784 题目大意:依次输入n个数,每当输入奇数个数的时候,求出当前序列的中位数(排好序的中位数). 此题可用各种方法求解. 排序二叉树方法,每个结点保存以其为根的左右子树中数的个数.如果数据出的够严格,这种方法会被卡的,除非是通过动态调整维持树的高度较小. 排序二叉树的代码如下: #include <cstdio> using namespace std; #define N 20000 struct Node { int v;…

2015-07-16 问题简述: 动态求取中位数的问题,输入一串数字,每输入第奇数个数时求取这些数的中位数. 原题链接:http://poj.org/problem?id=3784 解题思路: 求取中位数的方法常常想到使用堆来实现:取一个大顶堆,一个小顶堆,使大顶堆的堆顶记录中位数,因此,要时刻保持大顶堆堆顶元素小于小顶堆堆顶元素,且大顶堆元素个数等于小顶堆元素个数或等于小顶堆元素个数加一. 以下有两种堆得实现方法: 一:直接使用STL中的函数(make_heap,push_heap,pop_h…

Description For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far. Input The first line of input contains…

Description Moscow is hosting a major international conference, which is attended by n scientists from different countries. Each of the scientists knows exactly one language. For convenience, we enumerate all languages of the world with integers from…

Running Median Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3282 Description For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read…

[题目链接] http://poj.org/problem?id=3784 [算法] 对顶堆算法 要求动态维护中位数,我们可以将1-M/2(向下取整)小的数放在大根堆中,M/2+1-M小的数放在小根堆中 每次插入元素时,先将插入元素与小根堆堆顶比较,如果比堆顶小,则插入小根堆,否则,插入大根堆,然后,判断两个堆 的元素个数是否平衡,若不平衡,则交换两个堆的堆顶 [代码] #include <algorithm> #include <bitset> #include <ccty…

本题使用对顶堆做法. 为了动态维护中位数,我们可以建立两个堆 :一个大根对,一个小根堆. 用法:在动态维护的过程中,设当前的长度为length,大根堆存从小到大排名 $1 \thicksim \dfrac{m}{2} $ 的整数,小根堆存小到大排名 $ \dfrac{m}{2} + 1 \thicksim m $ 的整数 如何动态维护?顾名思义,动态,即边输入边处理.显然,为了维护中位数,我们还要不断地维护两个堆的\(size\) 每次新读入一个值,就 \(\begin{cases}插入大根堆&…

浅谈堆:https://www.cnblogs.com/AKMer/p/10284629.html 题目传送门:http://poj.org/problem?id=3784 用一个"对顶堆"动态维护中位数. 一个大根堆维护前半部分的权值,一个小根堆维护后半部分的权值. 新进来一个数如果小于大根对的权值就加进大根对,否则就加进小根堆. 每次动态维护大小,使得大根堆的大小为数字的一半. 大根对的堆顶就是中位数. 为了方便我把大根堆里的数取了个反也就变小根堆了. 时间复杂度:\(O(Tnlo…

Running Median Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3406   Accepted: 1576 Description For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output…