﻿<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:trackback="http://madskills.com/public/xml/rss/module/trackback/" xmlns:wfw="http://wellformedweb.org/CommentAPI/" xmlns:slash="http://purl.org/rss/1.0/modules/slash/"><channel><title>博客园-火火-随笔分类-c与数据结构</title><link>http://www.cnblogs.com/enuosky/category/57233.html</link><description>我快乐地努力着……，有时，也努力地快乐着</description><language>zh-cn</language><lastBuildDate>Sun, 27 Jul 2008 04:09:07 GMT</lastBuildDate><pubDate>Sun, 27 Jul 2008 04:09:07 GMT</pubDate><ttl>60</ttl><item><title>循环赛日程表算法</title><link>http://www.cnblogs.com/enuosky/archive/2008/07/27/1252377.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Sun, 27 Jul 2008 03:39:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2008/07/27/1252377.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/1252377.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2008/07/27/1252377.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/1252377.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/1252377.html</trackback:ping><description><![CDATA[摘要: 题目：有n=2^k个运动员要进行循环赛。现要设计一个满足以下要求的比赛日程表：&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;（1）每个选手必须与其他n-1个选手各赛一次...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2008/07/27/1252377.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/1252377.html?type=1" width = "1" height = "1" />]]></description></item><item><title>猴子称大王问题算法</title><link>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244589.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Wed, 16 Jul 2008 10:19:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244589.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/1244589.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244589.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/1244589.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/1244589.html</trackback:ping><description><![CDATA[摘要: 题目：一堆猴子都有编号，编号是1，2，3...m,这群猴子（m个）按照1-m的顺序围坐一圈，从第1开始数，每数到第N个，该猴子就要离开此圈，这样依次下来，直到圈中只剩下最后一只猴子，则该猴子为大王...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2008/07/16/1244589.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/1244589.html?type=1" width = "1" height = "1" />]]></description></item><item><title>自上至下，自左而右打印一棵二叉树(Tree)  算法</title><link>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244562.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Wed, 16 Jul 2008 09:36:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244562.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/1244562.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2008/07/16/1244562.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/1244562.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/1244562.html</trackback:ping><description><![CDATA[摘要: 自上至下，自左而右打印一棵二叉树(Tree)算法解题思路：采用两个队列（Queue）辅助实现。算法描述：Tree*t;//定义一棵二叉树，指向二叉树的根节点Queue*q1,*...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2008/07/16/1244562.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/1244562.html?type=1" width = "1" height = "1" />]]></description></item><item><title>再议公交查询算法</title><link>http://www.cnblogs.com/enuosky/archive/2007/01/24/629378.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Wed, 24 Jan 2007 08:46:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2007/01/24/629378.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/629378.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2007/01/24/629378.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/629378.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/629378.html</trackback:ping><description><![CDATA[摘要: &nbsp;&nbsp;&nbsp;前段时间写过一篇《公交路线查询算法》，其中设计了一个数据存储的方案，这里又做了一番改进。《公交路线查询算法》提到的算法最多提供倒乘一次的方案（我觉得在实际应用...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2007/01/24/629378.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/629378.html?type=1" width = "1" height = "1" />]]></description></item><item><title>公交路线查询算法</title><link>http://www.cnblogs.com/enuosky/archive/2006/12/30/607950.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Sat, 30 Dec 2006 07:55:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2006/12/30/607950.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/607950.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2006/12/30/607950.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/607950.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/607950.html</trackback:ping><description><![CDATA[摘要: 今天跟亮猪谈到公交查询，不负责任的说了句很简单吗&#8230;&#8230;但又一想不那么简单。费了点时间，想个如此算法：&nbsp;数据库中录入每一路公交走的线路包括途径站点，及站点间距...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2006/12/30/607950.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/607950.html?type=1" width = "1" height = "1" />]]></description></item><item><title>c程序与数据结构随笔整理</title><link>http://www.cnblogs.com/enuosky/archive/2006/05/08/394328.html</link><dc:creator>火火</dc:creator><author>火火</author><pubDate>Mon, 08 May 2006 13:30:00 GMT</pubDate><guid>http://www.cnblogs.com/enuosky/archive/2006/05/08/394328.html</guid><wfw:comment>http://www.cnblogs.com/enuosky/comments/394328.html</wfw:comment><comments>http://www.cnblogs.com/enuosky/archive/2006/05/08/394328.html#Feedback</comments><slash:comments>12</slash:comments><wfw:commentRss>http://www.cnblogs.com/enuosky/comments/commentRss/394328.html</wfw:commentRss><trackback:ping>http://www.cnblogs.com/enuosky/services/trackbacks/394328.html</trackback:ping><description><![CDATA[摘要: 以前，随笔没有分类，有些东西查阅起来就不那么方便了，现在整理一下。c程序（大多数是是课程设计时写的）：判断是否回文字符串(02-1517:05)单链表的交并差（c语言数据...&nbsp;&nbsp;<a href='http://www.cnblogs.com/enuosky/archive/2006/05/08/394328.html'>阅读全文</a><img src ="http://www.cnblogs.com/enuosky/aggbug/394328.html?type=1" width = "1" height = "1" />]]></description></item></channel></rss>