﻿<?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>C++博客-Coder Space-文章分类-数据结构</title><link>http://www.cppblog.com/Davidlrzh/category/13980.html</link><description /><language>zh-cn</language><lastBuildDate>Fri, 11 Jun 2010 08:58:35 GMT</lastBuildDate><pubDate>Fri, 11 Jun 2010 08:58:35 GMT</pubDate><ttl>60</ttl><item><title>PKU 2001 Shortest Prefixes --- 最短唯一前缀，Trie字典树</title><link>http://www.cppblog.com/Davidlrzh/articles/117565.html</link><dc:creator>David Liu</dc:creator><author>David Liu</author><pubDate>Thu, 10 Jun 2010 08:09:00 GMT</pubDate><guid>http://www.cppblog.com/Davidlrzh/articles/117565.html</guid><wfw:comment>http://www.cppblog.com/Davidlrzh/comments/117565.html</wfw:comment><comments>http://www.cppblog.com/Davidlrzh/articles/117565.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Davidlrzh/comments/commentRss/117565.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Davidlrzh/services/trackbacks/117565.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 题意：对给定的字典单词，求每个单词不具有二义性的最短前缀。解法：根据单词表，建立字典树，同时记录到达每个节点的单词数，单词数为1，即为最短前缀。最简单的字典树应用。源代码Code highlighting produced by Actipro CodeHighlighter (freeware)http://www.CodeHighlighter.com/-->&nbsp;&nbsp;1#i...&nbsp;&nbsp;<a href='http://www.cppblog.com/Davidlrzh/articles/117565.html'>阅读全文</a><img src ="http://www.cppblog.com/Davidlrzh/aggbug/117565.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Davidlrzh/" target="_blank">David Liu</a> 2010-06-10 16:09 <a href="http://www.cppblog.com/Davidlrzh/articles/117565.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>PKU 1451 T9 --- 手机输入法，Trie字典树+DFS</title><link>http://www.cppblog.com/Davidlrzh/articles/117271.html</link><dc:creator>David Liu</dc:creator><author>David Liu</author><pubDate>Sun, 06 Jun 2010 08:05:00 GMT</pubDate><guid>http://www.cppblog.com/Davidlrzh/articles/117271.html</guid><wfw:comment>http://www.cppblog.com/Davidlrzh/comments/117271.html</wfw:comment><comments>http://www.cppblog.com/Davidlrzh/articles/117271.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Davidlrzh/comments/commentRss/117271.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Davidlrzh/services/trackbacks/117271.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 题意：T9输入法，要求根据给出的字典，对给定的数字顺序，选择具有最高频率的词前缀。解法：经典的字典树问题。根据字典词，建立字典树(Trie树），节点保存其权值(频率），对给定的数字进行搜索(DFS)，确定最大值。&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 具有相同最大值是，输出字典序靠前前缀。源代码C...&nbsp;&nbsp;<a href='http://www.cppblog.com/Davidlrzh/articles/117271.html'>阅读全文</a><img src ="http://www.cppblog.com/Davidlrzh/aggbug/117271.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Davidlrzh/" target="_blank">David Liu</a> 2010-06-06 16:05 <a href="http://www.cppblog.com/Davidlrzh/articles/117271.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>PKU 3562 “Roman” corridor --- 罗马数字问题，Trie字典树+DFS</title><link>http://www.cppblog.com/Davidlrzh/articles/117225.html</link><dc:creator>David Liu</dc:creator><author>David Liu</author><pubDate>Sat, 05 Jun 2010 08:24:00 GMT</pubDate><guid>http://www.cppblog.com/Davidlrzh/articles/117225.html</guid><wfw:comment>http://www.cppblog.com/Davidlrzh/comments/117225.html</wfw:comment><comments>http://www.cppblog.com/Davidlrzh/articles/117225.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Davidlrzh/comments/commentRss/117225.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Davidlrzh/services/trackbacks/117225.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: &nbsp;题意：根据罗马数字与阿拉伯数的对应，在给定表格中，从左边走到右边（不能向左退），相邻格的走，到达右边后，求走过的罗马数字中的&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 合法串中的最小值。解法：先通过预处理，建立Trie字典树，利用字典树判断路径串的合法性及其值大小。Trie数据结构...&nbsp;&nbsp;<a href='http://www.cppblog.com/Davidlrzh/articles/117225.html'>阅读全文</a><img src ="http://www.cppblog.com/Davidlrzh/aggbug/117225.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Davidlrzh/" target="_blank">David Liu</a> 2010-06-05 16:24 <a href="http://www.cppblog.com/Davidlrzh/articles/117225.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>