﻿<?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-&lt;FONT face="Arial Black" size=12&gt;&lt;FONT color=#663399 qq&gt;F&lt;/FONT&gt;&lt;FONT color=#00ade5 qq&gt;e&lt;/FONT&gt;&lt;FONT color=#663399 qq&gt;l&lt;/FONT&gt;&lt;FONT color=#00b085 qq&gt;i&lt;/FONT&gt;&lt;FONT color=#cc6633 qq&gt;c&lt;/FONT&gt;&lt;FONT color=#00b085 qq&gt;i&lt;/FONT&gt;&lt;FONT color=#00ade5 qq&gt;a&lt;/FONT&gt;&lt;/FONT&gt;-随笔分类-动态规划</title><link>http://www.cppblog.com/Felicia/category/4908.html</link><description /><language>zh-cn</language><lastBuildDate>Mon, 19 May 2008 13:28:44 GMT</lastBuildDate><pubDate>Mon, 19 May 2008 13:28:44 GMT</pubDate><ttl>60</ttl><item><title>[动态规划]O(n^2 / logn)的LCS</title><link>http://www.cppblog.com/Felicia/archive/2007/10/19/34618.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Fri, 19 Oct 2007 08:56:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/19/34618.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/34618.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/19/34618.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/34618.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/34618.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 上次说，LCS有O(n^2 / logn)的解法。这个解法是在字符集不大的情况下，先预处理，再用位运算做状态转移。<br>唐文斌曾经翻译过一篇论文，专门讨论这个问题。<br><br>下面是练习题（n = 10000 的LCS）<br>http://acm.whu.edu.cn/oak/problem/problem.jsp?problem_id=1210<br><br>和我的解答<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/19/34618.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/34618.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-19 16:56 <a href="http://www.cppblog.com/Felicia/archive/2007/10/19/34618.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划] pku1458 最长公共子序列</title><link>http://www.cppblog.com/Felicia/archive/2007/10/16/34380.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 16 Oct 2007 14:46:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/16/34380.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/34380.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/16/34380.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/34380.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/34380.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 最长公共子序列……想必很多人都知道吧……<br>这里给出一个O(n^2)的算法，人人都会的。<br>但是，我想说，我所知道的最好算法，是O(n^2 / logn)的。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/16/34380.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/34380.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-16 22:46 <a href="http://www.cppblog.com/Felicia/archive/2007/10/16/34380.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1080</title><link>http://www.cppblog.com/Felicia/archive/2007/10/12/34069.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Fri, 12 Oct 2007 14:25:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/12/34069.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/34069.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/12/34069.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/34069.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/34069.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 很简单的DP，也是很基础的DP。做法就不说啦：）<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/12/34069.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/34069.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-12 22:25 <a href="http://www.cppblog.com/Felicia/archive/2007/10/12/34069.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1338</title><link>http://www.cppblog.com/Felicia/archive/2007/10/09/33851.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 09 Oct 2007 13:53:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/09/33851.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33851.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/09/33851.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33851.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33851.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 非常经典的递推计算。基本思想是设3个指针，分别表示3个素数乘到哪了，然后通过比较3个指针位置的递推结果来确定下一个数是什么。<br>具体实现见代码。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/09/33851.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33851.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-09 21:53 <a href="http://www.cppblog.com/Felicia/archive/2007/10/09/33851.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku3420</title><link>http://www.cppblog.com/Felicia/archive/2007/10/08/33740.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Mon, 08 Oct 2007 01:19:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/08/33740.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33740.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/08/33740.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33740.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33740.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典题型。如果列数较少，就能用我们熟知的状态压缩DP解决。但现在列数有2^31。考虑到相邻两列之间状态转移规则是相同的，我们可以用矩阵表示这种转移规则，而最后的结果就是求这个转移矩阵的n次幂的左上角元素。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/08/33740.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33740.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-08 09:19 <a href="http://www.cppblog.com/Felicia/archive/2007/10/08/33740.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1191</title><link>http://www.cppblog.com/Felicia/archive/2007/10/08/33737.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Mon, 08 Oct 2007 01:12:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/08/33737.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33737.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/08/33737.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33737.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33737.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 不错的DP题。状态f[i][x1][y1][x2][y2]表示要把(x1,y1) -- (x2, y2) 分割成i块所得到的最小平方和（平方和指的是每块矩形的和的平方和）。然后根据水平和竖直切割进行状态转移。这样计算出f[n][1][1][8][8]得到整个棋盘分割成n块得到的最小平方和，然后代入均方差公式算得结果。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/08/33737.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33737.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-08 09:12 <a href="http://www.cppblog.com/Felicia/archive/2007/10/08/33737.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1179</title><link>http://www.cppblog.com/Felicia/archive/2007/10/05/33515.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Fri, 05 Oct 2007 08:47:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/05/33515.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33515.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/05/33515.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33515.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33515.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典的DP，把环断开，f[i][j][0]记录i到j的最小值，f[i][j][1]记录最大值，然后递推计算。记录最小值是因为两个负数乘起来可能得到一个大的正数。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/05/33515.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33515.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-05 16:47 <a href="http://www.cppblog.com/Felicia/archive/2007/10/05/33515.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1189</title><link>http://www.cppblog.com/Felicia/archive/2007/10/04/33440.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Thu, 04 Oct 2007 12:47:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/10/04/33440.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33440.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/10/04/33440.html#Feedback</comments><slash:comments>4</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33440.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33440.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 概率+DP，比较经典的题。按照递推的方式计算概率。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/10/04/33440.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33440.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-10-04 20:47 <a href="http://www.cppblog.com/Felicia/archive/2007/10/04/33440.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1185</title><link>http://www.cppblog.com/Felicia/archive/2007/09/30/33271.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sun, 30 Sep 2007 14:09:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/30/33271.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33271.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/30/33271.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33271.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33271.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典的状态压缩DP，状态是f[i][j]，表示第i行，以3进制j为状态。j的位代表一个格子，只能是：0表示第i行和第i - 1行都没有炮兵，1表示第i行没有炮兵而第i-1行有炮兵，2表示第i行有炮兵。然后用DFS进行状态转移。一开始我做了超时，后来预处理了一下合法状态，快了不少，才AC。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/30/33271.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33271.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-30 22:09 <a href="http://www.cppblog.com/Felicia/archive/2007/09/30/33271.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1163</title><link>http://www.cppblog.com/Felicia/archive/2007/09/29/33228.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sat, 29 Sep 2007 14:43:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/29/33228.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33228.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/29/33228.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33228.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33228.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 今天郁闷了，贴个小代码<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/29/33228.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33228.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-29 22:43 <a href="http://www.cppblog.com/Felicia/archive/2007/09/29/33228.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]动态规划总结 by Amber</title><link>http://www.cppblog.com/Felicia/archive/2007/09/29/33222.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sat, 29 Sep 2007 12:25:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/29/33222.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/33222.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/29/33222.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/33222.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/33222.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 动态规划总结 by Amber<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/29/33222.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/33222.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-29 20:25 <a href="http://www.cppblog.com/Felicia/archive/2007/09/29/33222.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku3133</title><link>http://www.cppblog.com/Felicia/archive/2007/09/24/32806.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Mon, 24 Sep 2007 13:12:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/24/32806.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/32806.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/24/32806.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/32806.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/32806.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 强烈推荐此题！<br>状态压缩DP的好题！<br>分析见内<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/24/32806.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/32806.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-24 21:12 <a href="http://www.cppblog.com/Felicia/archive/2007/09/24/32806.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1038</title><link>http://www.cppblog.com/Felicia/archive/2007/09/12/32060.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Wed, 12 Sep 2007 12:44:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/12/32060.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/32060.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/12/32060.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/32060.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/32060.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典的状态压缩DP，《算法艺术与信息学竞赛》的例题。f[i][j]表示前i行，最后两行状态为二进制数j，嵌入的最多芯片数。第i行到第i+1行用DFS进行状态转移。<br>由于第i+1行只和第i行有关，故可以用滚动数组优化。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/12/32060.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/32060.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-12 20:44 <a href="http://www.cppblog.com/Felicia/archive/2007/09/12/32060.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku3375</title><link>http://www.cppblog.com/Felicia/archive/2007/09/11/32030.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 11 Sep 2007 14:28:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/11/32030.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/32030.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/11/32030.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/32030.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/32030.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: A O(NM) dynamic programming algorithm is quite apparent after sorting the computers and network interfaces by their coordinates. Furthermore, in any optimized case, for each computer the difference between the the indices of the network interfaces matching to and closest to the computer is never larger than N. So the complexity could be reduced to O(N2)<br><br>有很多细节不好考虑，应该是我的水平原因。最后我向updog要了数据才过的。而且代码写的不好。将就看一下吧。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/11/32030.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/32030.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-11 22:28 <a href="http://www.cppblog.com/Felicia/archive/2007/09/11/32030.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1170</title><link>http://www.cppblog.com/Felicia/archive/2007/09/04/31516.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 04 Sep 2007 00:37:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/04/31516.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31516.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/04/31516.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31516.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31516.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 呼~今天去学校啦！早上7点起床写题，挑了个简单题写 ^_^<br>这个是IOI95的DP题。用一个b位的6进制数i表示状态。这个6进制数的每一位分别表示相应物品的数量。f[i]表示状态i下的最小花费。同样也可以用6进制数j表示优惠。那么，f[i]就能转移到f[i - j]，如果优惠j可用的话。代价是使用优惠j时减少的花费。最后的答案就是min(f[i])，0 <= i <= start（start是初始状态）。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/04/31516.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31516.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-04 08:37 <a href="http://www.cppblog.com/Felicia/archive/2007/09/04/31516.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1160</title><link>http://www.cppblog.com/Felicia/archive/2007/09/03/31493.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Mon, 03 Sep 2007 14:44:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/03/31493.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31493.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/03/31493.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31493.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31493.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 先预处理，把第i个村子到第j个村子中，建一个邮局的最小代价算出来，存在min_cost[i][j]里。<br>接下来就可以DP。设f[i][j]为前i个邮局，建在前j个村子的最小代价。那么f[i][j]可以转移到f[i + 1][j + k]，(1 <= k 且 j + k <= n)，代价是min_cost[j + 1][j + k]。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/03/31493.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31493.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-03 22:44 <a href="http://www.cppblog.com/Felicia/archive/2007/09/03/31493.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1088</title><link>http://www.cppblog.com/Felicia/archive/2007/09/02/31411.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sun, 02 Sep 2007 12:02:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/02/31411.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31411.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/02/31411.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31411.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31411.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 简单的记忆化搜索。很早以前做的，代码风格很乱。将就一下啦。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/02/31411.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31411.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-02 20:02 <a href="http://www.cppblog.com/Felicia/archive/2007/09/02/31411.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1737</title><link>http://www.cppblog.com/Felicia/archive/2007/09/02/31386.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sun, 02 Sep 2007 05:53:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/02/31386.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31386.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/02/31386.html#Feedback</comments><slash:comments>6</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31386.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31386.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 楼爷的题。递推。f[n]表示n个结点的连通图个数，则有递推公式：<br><br>void calc(int n)<br>{<br>    f[n] = 0;<br>    for (int i = 1; i < n; i++)<br>        f[n] += f[i] * f[n - i] * (pow(i) - 1) * C(n - 2, i - 1);<br>    //pow(x) == 2^x<br>}<br><br>因为数据较多，所以预先算出f[1] -- f[50]，再输出。要用高精度。我用了标程。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/02/31386.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31386.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-02 13:53 <a href="http://www.cppblog.com/Felicia/archive/2007/09/02/31386.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1946</title><link>http://www.cppblog.com/Felicia/archive/2007/09/01/31353.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sat, 01 Sep 2007 03:42:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/09/01/31353.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31353.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/09/01/31353.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31353.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31353.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 首先明确一点：最优解必为奶牛1..n-1轮流领跑，奶牛n撞线。且跑了x圈后，未领跑过的奶牛都耗费了x的体力。<br>设f[i][j][k]表示前i-1头奶牛已领跑，现在由第i头奶牛领跑，一共跑了j圈，奶牛i耗费了k的体力。<br>则f[i][j][k]可以转移到f[i][j + p][k + p^2]（耗费1分钟，奶牛i以p圈/分钟的速度继续领跑），也可转移到f[i + 1][j][j]（换成奶牛i + 1领跑，不耗费时间）。<br>时间复杂度为O(nde^2.5)。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/09/01/31353.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31353.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-09-01 11:42 <a href="http://www.cppblog.com/Felicia/archive/2007/09/01/31353.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1947</title><link>http://www.cppblog.com/Felicia/archive/2007/08/31/31321.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Fri, 31 Aug 2007 10:27:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/31/31321.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31321.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/31/31321.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31321.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31321.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 推荐此题。基础树型DP。<br>f[x][i](1 <= i <= p)表示以x为根的子树，变成剩下i个点的子树，且剩余子树包含根结点，需要去掉的最少边数。<br>那么父结点的f值可以由它所有的儿子的f值做背包得到。<br>最后的答案是min(min(f[i][p]) + 1 (2 <= i <= n), f[1][p])<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/31/31321.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31321.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-31 18:27 <a href="http://www.cppblog.com/Felicia/archive/2007/08/31/31321.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1848</title><link>http://www.cppblog.com/Felicia/archive/2007/08/30/31236.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Thu, 30 Aug 2007 13:47:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/30/31236.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31236.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/30/31236.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31236.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31236.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 强烈推荐此题。树型DP。<br>分析较长且带有图示，请阅读全文。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/30/31236.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31236.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-30 21:47 <a href="http://www.cppblog.com/Felicia/archive/2007/08/30/31236.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1112</title><link>http://www.cppblog.com/Felicia/archive/2007/08/29/31169.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Wed, 29 Aug 2007 09:15:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/29/31169.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31169.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/29/31169.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31169.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31169.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 强烈推荐此题。图论和DP结合。<br>分析较长，请阅读全文。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/29/31169.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31169.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-29 17:15 <a href="http://www.cppblog.com/Felicia/archive/2007/08/29/31169.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku2411</title><link>http://www.cppblog.com/Felicia/archive/2007/08/28/31052.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 28 Aug 2007 13:03:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/28/31052.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31052.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/28/31052.html#Feedback</comments><slash:comments>12</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31052.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31052.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典的状态压缩DP。f[i][j]表示第i行，方格排布为二进制数j（第k位上为1表示凸出一个格子，为0表示不凸出）的方案数。用DFS进行状态转移。<br>如果行数比较多的话，可以用矩阵乘法优化。因为每行的状态转移都是相同的。设烈数为m，行数为n，可以做到O(2^(3m)logn)。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/28/31052.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31052.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-28 21:03 <a href="http://www.cppblog.com/Felicia/archive/2007/08/28/31052.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku2288</title><link>http://www.cppblog.com/Felicia/archive/2007/08/28/31050.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Tue, 28 Aug 2007 12:47:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/28/31050.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/31050.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/28/31050.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/31050.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/31050.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典的TSP问题变种。状态为f[i][j][k]，表示经过二进制数i所指的哈密顿路（第bi位为1表示经过该点，为0表示不经过该点），倒数第二个点为j，最后一个点为k。.value表示最大权值，.num表示能走出最大权值的路径数。若图中k到p有边，f[i][j][k]则转移到f[i'][k][p]。i' == i | (1 << p)。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/28/31050.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/31050.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-28 20:47 <a href="http://www.cppblog.com/Felicia/archive/2007/08/28/31050.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1141</title><link>http://www.cppblog.com/Felicia/archive/2007/08/27/30936.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Mon, 27 Aug 2007 07:55:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/27/30936.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/30936.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/27/30936.html#Feedback</comments><slash:comments>1</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/30936.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/30936.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: int f[i][j]表示第i个字符到第j个字符需要添加的最少括号数。string ans[i][j] 表示第i个字符到第j个字符按照最优方案添加括号后的串。状态转移：1.f[i][j]由f[i + 1][j - 1]转移来（通过两端添括号() / [] ）。2.f[i][j]由f[i][k] + f[k + 1][j]转移来（通过串合并）。答案是ans[0][len - 1]。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/27/30936.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/30936.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-27 15:55 <a href="http://www.cppblog.com/Felicia/archive/2007/08/27/30936.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku1050</title><link>http://www.cppblog.com/Felicia/archive/2007/08/26/30849.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sun, 26 Aug 2007 05:51:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/26/30849.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/30849.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/26/30849.html#Feedback</comments><slash:comments>6</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/30849.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/30849.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 枚举矩形的上边和下边，花费O(n^2)，把问题转化成一维的最大M子段和，做一个O(n)的DP。<br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/26/30849.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/30849.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-26 13:51 <a href="http://www.cppblog.com/Felicia/archive/2007/08/26/30849.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku 部分动态规划题目列表</title><link>http://www.cppblog.com/Felicia/archive/2007/08/26/30848.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Sun, 26 Aug 2007 03:52:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/26/30848.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/30848.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/26/30848.html#Feedback</comments><slash:comments>3</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/30848.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/30848.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: pku 部分动态规划题目列表<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/26/30848.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/30848.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-26 11:52 <a href="http://www.cppblog.com/Felicia/archive/2007/08/26/30848.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>[动态规划]pku3345</title><link>http://www.cppblog.com/Felicia/archive/2007/08/15/30096.html</link><dc:creator>Felicia</dc:creator><author>Felicia</author><pubDate>Wed, 15 Aug 2007 10:42:00 GMT</pubDate><guid>http://www.cppblog.com/Felicia/archive/2007/08/15/30096.html</guid><wfw:comment>http://www.cppblog.com/Felicia/comments/30096.html</wfw:comment><comments>http://www.cppblog.com/Felicia/archive/2007/08/15/30096.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Felicia/comments/commentRss/30096.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Felicia/services/trackbacks/30096.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 建立一个虚点（权为无穷大），从它到每个入度为 0 的点都连一条边，然后做树型DP。<br>先递归算出子结点的 f 值，然后用背包的方法计算父结点的 f 值。<br><br>&nbsp;&nbsp;<a href='http://www.cppblog.com/Felicia/archive/2007/08/15/30096.html'>阅读全文</a><img src ="http://www.cppblog.com/Felicia/aggbug/30096.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Felicia/" target="_blank">Felicia</a> 2007-08-15 18:42 <a href="http://www.cppblog.com/Felicia/archive/2007/08/15/30096.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>