POJ3784:Running Median】的更多相关文章

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…
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…
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…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3282 Running Median 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 th…
1.cmd窗口中输入emulator -avd test 启动AVD时报错: Running an x86 based Android Virtual Device (AVD) is 10x faster. We strongly recommend creating a new AVD. 这个说的是使用一个x86的模拟器比使用arm内核的要快10倍,所以这只是个提示,让你换成内核为x86的模拟器呢!不换的话你就只能等了,操作每一步都可能要等个好三四秒. 我的是等了6.7分钟界面一直卡着. 2.…
线程池有5种状态:Running, SHUTDOWN, STOP, TIDYING, TERMINATED. 线程池状态定义代码如下: private final AtomicInteger ctl = new AtomicInteger(ctlOf(RUNNING, 0)); private static final int COUNT_BITS = Integer.SIZE - 3; private static final int CAPACITY = (1 << COUNT_BITS)…
Running Median 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…
Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1670   Accepted: 823 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 (midd…
由于我不会讲对顶堆,所以这里直接传上一个巨佬的学习笔记. 对顶堆其实还是很容易理解的,想这题的时候自己猜做法也能把没学过的对顶堆给想出来.后来了解,对顶堆主要还是动态的在线维护集合$K$大值.当然也可以带删除.但是可能退化,具体见另外一个题的说明. 像这题维护中位数就是要求第$N/2+1$大的数,所以可以让大根堆维护前$n/2$项,小根堆维护后$n/2+1$项(这里指排好序的).然后每次插入的时候及时调整即可,由于调整幅度不大,所以可以保证复杂度是$log$级别的. 代码被我写繁了因为我当时还不…
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 a single int…
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…
Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examples: [2,3,4] , the median is 3 [2,3], the median is (2 + 3) / 2 = 2.5 Design a d…
确保HADOOP_CONF_DIR或者YARN_CONF_DIR指向hadoop集群配置文件目录.这些配置用来写数据到hdfs以及连接yarn ResourceManager.(在$SPARK_HOME/conf/spark-env.sh中,添加export HADOOP_CONF_DIR=/home/koushengrui/app/hadoop/etc/hadoop).The configuration contained in this directory will be distribut…
一. 题目描写叙述 Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examples: [2,3,4] , the median is 3 [2,3], the median is (2 + 3) / 2 = 2.5…
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…
Running Apache Storm Securely Apache Storm offers a range of configuration options when trying to secure your cluster. By default all authentication and authorization is disabled but can be turned on as needed. Firewall/OS level Security You can stil…
代码链接:Tetris 码云地址 一.Alpha版本测试报告 1.测试计划 测试内容 1.方块生成下落 2.方块左右移动 3.方块快速下滑 4.本汉字替换 5.下一个汉字提示 6.方块颜色生成 7.汉字方块合并 8.游戏暂停和开始 9.背景音频 10.汉字合成分数变化 测试人员 陆海燕.蔡苑菲 测试时间 11.5-11.7 测试方法 android studio模拟器 安卓手机开启调试模式测试 质量目标 1-9功能正常,10功能附加 修复人员 马仪生 2.测试过程 测试截图 错误记录(提交iss…
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…
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;…
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&…
[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…
题目链接:http://ac.jobdu.com/problem.php?pid=1004 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1004 Median.cpp // Jobdu // // Created by PengFei_Zheng on 26/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. // #include <std…
4. Median of Two Sorted Arrays 官方的链接:4. Median of Two Sorted Arrays Description : There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n…
转载请注明出处:http://www.cnblogs.com/skywang12345/p/3509960.html 本章介绍线程池的生命周期.在"Java多线程系列--"基础篇"01之 基本概念"中,我们介绍过,线程有5种状态:新建状态,就绪状态,运行状态,阻塞状态,死亡状态.线程池也有5种状态:然而,线程池不同于线程,线程池的5种状态是:Running, SHUTDOWN, STOP, TIDYING, TERMINATED. 线程池状态定义代码如下: priv…
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题目链接: A. Median Smoothing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output A schoolboy named Vasya loves reading books on programming and mathematics. He has recently read an encyclopedia a…