﻿<?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>IT博客-苍月流影-最新评论</title><link>http://www.cnitblog.com/Acrux/CommentsRSS.aspx</link><description>冗余的代码</description><language>zh-cn</language><pubDate>Tue, 04 Mar 2008 13:29:03 GMT</pubDate><lastBuildDate>Tue, 04 Mar 2008 13:29:03 GMT</lastBuildDate><generator>cnblogs</generator><item><title>re: ZOJ 1505 Solitaire</title><link>http://www.cnitblog.com/Acrux/archive/2008/05/08/40171.html#43491</link><dc:creator>x</dc:creator><author>x</author><pubDate>Wed, 07 May 2008 16:08:00 GMT</pubDate><guid>http://www.cnitblog.com/Acrux/archive/2008/05/08/40171.html#43491</guid><description><![CDATA[我想问一下，在每个bfs中，在第一次overlap ( tp, i ) 失败后会执行<br>tp.g[i].x += dir[j][0], tp.g[i].y += dir[j][1];<br><br>在执行完这句后应该还需要一次overlap ( tp, i )判断吧？因为跳过一个子可能还有一个子这样两个字就重合了，如果数据变态一点的话，maybe wa?。<br><br>是不是应该这样写：<br>            if ( overlap ( tp, i ) )<br>                     tp.g[i].x += dir[j][0], tp.g[i].y += dir[j][1];<br>                 if ( !check ( tp.g[i].x, tp.g[i].y ) )<br>                     continue;<br>                 if ( overlap ( tp, i ) )<br>                    continue;<br><img src ="http://www.cnitblog.com/Acrux/aggbug/43491.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cnitblog.com/Acrux/" target="_blank">x</a> 2008-05-08 00:08 <a href="http://www.cnitblog.com/Acrux/archive/2008/05/08/40171.html#43491#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>