﻿<?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-pzz-文章分类-算法学习</title><link>http://www.cppblog.com/panzhizhou/category/19230.html</link><description /><language>zh-cn</language><lastBuildDate>Fri, 09 Aug 2013 16:07:46 GMT</lastBuildDate><pubDate>Fri, 09 Aug 2013 16:07:46 GMT</pubDate><ttl>60</ttl><item><title>动态规划-释放囚犯问题</title><link>http://www.cppblog.com/panzhizhou/articles/dp.html</link><dc:creator>pzz</dc:creator><author>pzz</author><pubDate>Wed, 07 Aug 2013 05:10:00 GMT</pubDate><guid>http://www.cppblog.com/panzhizhou/articles/dp.html</guid><wfw:comment>http://www.cppblog.com/panzhizhou/comments/202391.html</wfw:comment><comments>http://www.cppblog.com/panzhizhou/articles/dp.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/panzhizhou/comments/commentRss/202391.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/panzhizhou/services/trackbacks/202391.html</trackback:ping><description><![CDATA[挑战程序设计竞赛p47页，对于释放囚犯问题泛化为如果某一个问题的解，依赖于子问题的解，且没有重叠，那么就可以从底至上的迭代解决，通保存子问题的解来求得最钟问题的解。<br />这里我们看到对于当前释放的囚犯，那么知道当前最优解等于释放左边当中全部囚犯最优解+释放右边当中全部囚犯最优解+当前解（=A[j]-A[i]-2，考虑A[j],A[i]是两个端点）<br />这里我们注意初始化二维矩阵，最终可以求得一个上三角矩阵。<br /><br /><img src ="http://www.cppblog.com/panzhizhou/aggbug/202391.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/panzhizhou/" target="_blank">pzz</a> 2013-08-07 13:10 <a href="http://www.cppblog.com/panzhizhou/articles/dp.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>