﻿<?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-大成至圣-随笔分类-背包问题</title><link>http://www.cppblog.com/yaobing/category/10122.html</link><description /><language>zh-cn</language><lastBuildDate>Tue, 07 Apr 2009 13:49:06 GMT</lastBuildDate><pubDate>Tue, 07 Apr 2009 13:49:06 GMT</pubDate><ttl>60</ttl><item><title>P02: 完全背包问题</title><link>http://www.cppblog.com/yaobing/archive/2009/04/07/79205.html</link><dc:creator>姚冰</dc:creator><author>姚冰</author><pubDate>Tue, 07 Apr 2009 12:16:00 GMT</pubDate><guid>http://www.cppblog.com/yaobing/archive/2009/04/07/79205.html</guid><wfw:comment>http://www.cppblog.com/yaobing/comments/79205.html</wfw:comment><comments>http://www.cppblog.com/yaobing/archive/2009/04/07/79205.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/yaobing/comments/commentRss/79205.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/yaobing/services/trackbacks/79205.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 题目有N种物品和一个容量为V的背包，每种物品都有无限件可用。第i种物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使这些物品的费用总和不超过背包容量，且价值总和最大。基本思路这个问题非常类似于01背包问题，所不同的是每种物品有无限件。也就是从每种物品的角度考虑，与它相关的策略已并非取或不取两种，而是有取0件、取1件、取2件&#8230;&#8230;等很多种。如果仍然按照...&nbsp;&nbsp;<a href='http://www.cppblog.com/yaobing/archive/2009/04/07/79205.html'>阅读全文</a><img src ="http://www.cppblog.com/yaobing/aggbug/79205.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/yaobing/" target="_blank">姚冰</a> 2009-04-07 20:16 <a href="http://www.cppblog.com/yaobing/archive/2009/04/07/79205.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>P01:01背包问题</title><link>http://www.cppblog.com/yaobing/archive/2009/04/06/79077.html</link><dc:creator>姚冰</dc:creator><author>姚冰</author><pubDate>Mon, 06 Apr 2009 03:43:00 GMT</pubDate><guid>http://www.cppblog.com/yaobing/archive/2009/04/06/79077.html</guid><wfw:comment>http://www.cppblog.com/yaobing/comments/79077.html</wfw:comment><comments>http://www.cppblog.com/yaobing/archive/2009/04/06/79077.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/yaobing/comments/commentRss/79077.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/yaobing/services/trackbacks/79077.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 题目有N件物品和一个容量为V的背包。第i件物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。基本思路这是最基础的背包问题，特点是：每种物品仅有一件，可以选择放或不放。用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：f[i][v]=max{f[i-1][v],f[i-1][v-c[i...&nbsp;&nbsp;<a href='http://www.cppblog.com/yaobing/archive/2009/04/06/79077.html'>阅读全文</a><img src ="http://www.cppblog.com/yaobing/aggbug/79077.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/yaobing/" target="_blank">姚冰</a> 2009-04-06 11:43 <a href="http://www.cppblog.com/yaobing/archive/2009/04/06/79077.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>