<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/alex0</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; alex0]]></copyright><item><title><![CDATA[brogue 正则 搜索非Seed非food的内容来找出食物多的楼层 环视]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/11856273</link><guid>https://blog.youkuaiyun.com/alex0/article/details/11856273</guid><author>alex0</author><pubDate>Fri, 20 Sep 2013 22:45:59 +0800</pubDate><description><![CDATA[^((?!(Seed|food)).)+$


实际上用awk改进比较容易。正则本身不那么好算？]]></description><category></category></item><item><title><![CDATA[DP 背包问题 01背包]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/10166827</link><guid>https://blog.youkuaiyun.com/alex0/article/details/10166827</guid><author>alex0</author><pubDate>Thu, 22 Aug 2013 01:11:45 +0800</pubDate><description><![CDATA[DP看了十几次了 老忘
比较需要系统性的记一下。


以下是最简单的背包：
m为物品个数
n为背包体积
输入m个物品价值和体积，保存在v[]和w[]中
for i=1 to m
  for j=1 to n
    f[i][j] = max(f[i-1][j], f[i][j-v[i]] + w[i]) // 如果j-v[i]小于0就不考虑此项]]></description><category></category></item><item><title><![CDATA[plan]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/10087871</link><guid>https://blog.youkuaiyun.com/alex0/article/details/10087871</guid><author>alex0</author><pubDate>Mon, 19 Aug 2013 23:09:20 +0800</pubDate><description><![CDATA[glibc已基本看完
glib待看，加入翻译
stl待看]]></description><category></category></item><item><title><![CDATA[【python】如何下载一个网页]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9401577</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9401577</guid><author>alex0</author><pubDate>Sun, 21 Jul 2013 22:30:38 +0800</pubDate><description><![CDATA[Stackoverflow的答案：
http://stackoverflow.com/questions/22676/how-do-i-download-a-file-over-http-using-python
import urllib2
response = urllib2.urlopen('http://www.example.com/')
html = response.read()]]></description><category></category></item><item><title><![CDATA[OSPF RIP STP TRILL]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9394593</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9394593</guid><author>alex0</author><pubDate>Sun, 21 Jul 2013 00:41:02 +0800</pubDate><description><![CDATA[OSPF（开放式最短路径优先）隶属于IGP（内部网关协议），用于在单一自治系统内决策路由，是对链路状态路由协议的一种实现。
OSPF是SPF类路由协议中的开放式版本。
使用算法：dijkstra（分解视频：http://v.youku.com/v_show/id_XMjQyOTY1NDQw.html）
注意到OSPF是要用在同一个路由域中的，也指一个自治系统。所有域内路由维护同样数据的数据库]]></description><category></category></item><item><title><![CDATA[奇怪的C++ static]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9385675</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9385675</guid><author>alex0</author><pubDate>Fri, 19 Jul 2013 20:52:40 +0800</pubDate><description><![CDATA[原来函数和变量用static修饰没什么问题


在C++里有一些额外的static作用
在类中 static函数只能访问static成员（奇怪的玩法）
static变量无法在类中初始化


private的static变量可以在外部初始化（有且仅有一次）
const static int可以在类中初始化]]></description><category></category></item><item><title><![CDATA[知识图谱]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9385085</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9385085</guid><author>alex0</author><pubDate>Fri, 19 Jul 2013 19:26:30 +0800</pubDate><description><![CDATA[IOS - 无
ANDROID - 无
计算机网络 - 精通
LINUX - 精通
LINUX内核 - 熟悉
机器学习 - 擅长
SDN - 擅长
C - 精通
NOSQL - 了解
HADOOP - 了解]]></description><category></category></item><item><title><![CDATA[设计模式]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9337859</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9337859</guid><author>alex0</author><pubDate>Tue, 16 Jul 2013 00:43:36 +0800</pubDate><description><![CDATA[工厂模式：
提供接口，屏蔽实现，使用通用的一个factory进行实例的制造，内部的实例可变。工厂模式是针对一个产品的
抽象工厂模式：
针对多个产品，使用通用的]]></description><category></category></item><item><title><![CDATA[八皇后 & 迷宫 算法]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9322225</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9322225</guid><author>alex0</author><pubDate>Sun, 14 Jul 2013 01:00:36 +0800</pubDate><description><![CDATA[预留。


八皇后实际上可以递归实现，迷宫也是。
但递归都可以转成迭代。尽量思考吧。]]></description><category></category></item><item><title><![CDATA[qsort 快排算法讲解]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9321783</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9321783</guid><author>alex0</author><pubDate>Sat, 13 Jul 2013 23:27:52 +0800</pubDate><description><![CDATA[//快速排序
void quick_sort(int s[], int l, int r)
{
    if (l < r)
    {
              //Swap(s[l], s[(l + r) / 2]); //将中间的这个数和第一个数交换 参见注1
        int i = l, j = r, x = s[l];
        while (i < j)
       ]]></description><category></category></item><item><title><![CDATA[机器学习术语发音]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9321707</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9321707</guid><author>alex0</author><pubDate>Sat, 13 Jul 2013 23:10:57 +0800</pubDate><description><![CDATA[artificial neural network
美 [ˌɑrtəˈfɪʃəl ˈnʊrəl ˈnɛtˌwɚk]



genetic algorithm
美 [dʒəˈnɛtɪk ˈælɡəˌrɪðəm]]]></description><category></category></item><item><title><![CDATA[全排列算法]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9320933</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9320933</guid><author>alex0</author><pubDate>Sat, 13 Jul 2013 21:17:46 +0800</pubDate><description><![CDATA[图画的不错，是遍历得到全排列。
http://blog.youkuaiyun.com/joylnwang/article/details/7064115


也可以直接用“求下一个排列”的方法来做。STL有相关函数。
http://blog.youkuaiyun.com/yxjiang/article/details/3013574

http://plutoblog.iteye.com/blog/9762]]></description><category></category></item><item><title><![CDATA[MAP REDUCE小计]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9319145</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9319145</guid><author>alex0</author><pubDate>Sat, 13 Jul 2013 16:09:47 +0800</pubDate><description><![CDATA[MAP（映射）就是将需要做的任务分解，分派到多个计算节点上，让每个计算节点执行一个单独的子集
REDUCE（化简）就是将所有运算结果统筹到一起，这个过程会尽量在单个节点上完成，因为一般它是难以并行化的


NOSQL
丧失了ACID原则，遵从CAP理论，只能满足CAP的两者。一般可以和MAP REDUCE结合的比较好。]]></description><category></category></item><item><title><![CDATA[冒泡的最佳时间 O(n)]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9318003</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9318003</guid><author>alex0</author><pubDate>Sat, 13 Jul 2013 12:53:45 +0800</pubDate><description><![CDATA[http://www.cnblogs.com/melon-h/archive/2012/09/20/2694941.html
public void bubbleSort(int arr[]) {
    boolean didSwap;
    for(int i = 0, len = arr.length; i < len - 1; i++) {
        didSwap = fals]]></description><category></category></item><item><title><![CDATA[NOSQL学习]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9315655</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9315655</guid><author>alex0</author><pubDate>Fri, 12 Jul 2013 22:47:46 +0800</pubDate><description><![CDATA[数据模型

　　传统的数据库在数据模型方面，主要是关系型，它的特色是对Join类操作和ACID事务的支持。在NoSQL领域，主要有三种主流的数据模型：

　　Column-oriented(列式)

　　列式也主要使用Table这样的模型，但是它并不支持类似Join这样多表的操作，它的主要特点是在存储数据时，主要围绕着“列(Column)”，而不是像传统的关系型数据库那样根据“行(Ro]]></description><category></category></item><item><title><![CDATA[背包复习]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9305787</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9305787</guid><author>alex0</author><pubDate>Fri, 12 Jul 2013 00:59:57 +0800</pubDate><description><![CDATA[背包本质是动归，动归实际就是有条理的穷举。01背包就是物品只能放一次的背包，对应的函数在迭代时注意不能叠加物品。
常见的背包就是无穷多次背包，可以引用自己本身的函数（引用装的更少时候的状态）


http://blog.youkuaiyun.com/xiaowei_cqu/article/details/8191808]]></description><category></category></item><item><title><![CDATA[Cracking the coding interview--Q9]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9295905</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9295905</guid><author>alex0</author><pubDate>Thu, 11 Jul 2013 00:16:23 +0800</pubDate><description><![CDATA[Chapter 9 | Sorting and Searching

9.1 You are given two sorted arrays, A and B, and A has a large enough buffer at the end to hold B. Write a method to merge B into A in sorted order.

插入排序，需要调整数]]></description><category></category></item><item><title><![CDATA[Cracking the coding interview--Q5]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9295499</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9295499</guid><author>alex0</author><pubDate>Wed, 10 Jul 2013 23:06:35 +0800</pubDate><description><![CDATA[Chapter 5 | Bit Manipulation

5.1 You are given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e.g., M becomes a
 subst]]></description><category></category></item><item><title><![CDATA[Cracking the coding interview--Q4]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9294899</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9294899</guid><author>alex0</author><pubDate>Wed, 10 Jul 2013 22:33:30 +0800</pubDate><description><![CDATA[Chapter 4 | Trees and Graphs

4.1 Implement a function to check if a tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that no two leaf nodes differ
]]></description><category></category></item><item><title><![CDATA[Cracking the coding interview--Q3]]></title><link>https://blog.youkuaiyun.com/alex0/article/details/9286681</link><guid>https://blog.youkuaiyun.com/alex0/article/details/9286681</guid><author>alex0</author><pubDate>Wed, 10 Jul 2013 01:03:30 +0800</pubDate><description><![CDATA[Chapter 3 | Stacks and Queues

3.1 Describe how you could use a single array to implement three stacks.

分段呗- -

3.2 How would you design a stack which, in addition to push and pop, also has a f]]></description><category></category></item></channel></rss>