﻿<?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-王之昊的小屋-文章分类-pku</title><link>http://www.cppblog.com/Wangzhihao/category/15609.html</link><description>为了自己的梦想，每天要像小乐一样勤奋！</description><language>zh-cn</language><lastBuildDate>Mon, 03 Jan 2011 21:12:13 GMT</lastBuildDate><pubDate>Mon, 03 Jan 2011 21:12:13 GMT</pubDate><ttl>60</ttl><item><title>1444 Parallelepiped walk</title><link>http://www.cppblog.com/Wangzhihao/articles/120959.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Wed, 21 Jul 2010 06:29:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/120959.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/120959.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/120959.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/120959.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/120959.html</trackback:ping><description><![CDATA[

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<p style="margin-bottom: 0cm;" align="CENTER"><font face="AR PL UMing CN, serif">1444
Parallelepiped walk</font></p>
<p style="margin-bottom: 0cm;" align="LEFT"><a name="CategoryEntryList1_EntryStoryList_Entries_ctl01_TitleUrl"></a>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<font face="AR PL UMing CN, serif">	</font>做法一可以尝试分类讨论。具体类似这道题目<font face="AR PL UMing CN, serif"><a href="http://www.cppblog.com/Wangzhihao/archive/2010/06/23/118513.html">The
Return of Carl</a></font>。然后也可以观察分类讨论的几种情况。如下图：</p>
<p style="margin-bottom: 0cm;" align="LEFT"><img src="http://www.cppblog.com/images/cppblog_com/wangzhihao/sample.jpg" border="0"></p>
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</p><p style="margin-bottom: 0cm;" align="LEFT">可以发现他们沿<font face="AR PL UMing CN, serif">x</font>轴一个方向走，沿<font face="AR PL UMing CN, serif">y</font>轴也是一个方向走，不会来回。（这点很重要，如果不发现这一点直接搜的话，会搜出更短的但不合法的距离。）
也可以判是否路线经过了所有的矩形，如果经过了，就是一条合法路径。</p>
<p>&nbsp;</p><img src ="http://www.cppblog.com/Wangzhihao/aggbug/120959.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-07-21 14:29 <a href="http://www.cppblog.com/Wangzhihao/articles/120959.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Cover an Arc</title><link>http://www.cppblog.com/Wangzhihao/articles/119376.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Mon, 05 Jul 2010 12:19:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/119376.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/119376.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/119376.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/119376.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/119376.html</trackback:ping><description><![CDATA[

<div style="text-align: center;"><span style="font-size: 18pt;"></span><a href="http://acm.pku.edu.cn/JudgeOnline/problem?id=1266"><span style="font-size: 18pt;">Cover an Arc</span></a><br><span style="font-size: 18pt;"></span></div>
<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 写好计算几何很多时候要养成一个好习惯。<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 比如求比浮点数 f 小的最小的整数，认为绝对值差在1e-9的精度内两个浮点数是相同的，首先是看 f 是否就是一个整数， 是则返回 f 。 否则返回 f 的截断。 这时如果直接用 floor( f )，很容易忽视这种情况 如0.9，999，999，999。它和 1 在1e-9精度下是相等的。所以答案应该是1， 而floor(f) = 0;<br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 又如<a href="http://acm.pku.edu.cn/JudgeOnline/problem?id=1263">Reflections</a> 这道题中的反射，大致是光线从一点射出，射到某个球上，再反射，再射下一个球上，重复，直到不再与球相交。<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 我们假设前一次反射的球为base，再求下一个射到的球的位置时，就不该考虑base，他一定不是我们要求的那个下一个球，但如果不先排除base，我们很容易就算到base是下一个射到的球这样一个错误的答案，这样模棱两可的情况如果能在逻辑上排除，尽量排除。<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 在这些精度问题上要注意小心，更好的办法就是平时写的时候养成一个好习惯。<br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 能逻辑上排除的就不进行浮点数比较，能避免就避免。如果不能避免，一定要加上符号判断函数 <br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; int D(double x ){ return x &lt; -eps ? -1 : x &gt; eps; }<br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/119376.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-07-05 20:19 <a href="http://www.cppblog.com/Wangzhihao/articles/119376.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>2284 That Nice Euler Circuit</title><link>http://www.cppblog.com/Wangzhihao/articles/110299.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Mon, 22 Mar 2010 08:25:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/110299.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/110299.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/110299.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/110299.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/110299.html</trackback:ping><description><![CDATA[这个图形显然是一个平面性图,然后必定满足欧拉公式.F - V + E = 2; 接下来只要去求有多少个点,多少条边即可.<br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/110299.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-22 16:25 <a href="http://www.cppblog.com/Wangzhihao/articles/110299.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3329 Twirl Around</title><link>http://www.cppblog.com/Wangzhihao/articles/110187.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Sat, 20 Mar 2010 16:36:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/110187.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/110187.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/110187.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/110187.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/110187.html</trackback:ping><description><![CDATA[&nbsp; 这道题模拟，已知一个支点，去寻找下一个支点，累加这期间的角度直到转到满意的角度为止。把绕支点的转动看成一个圆，把多边形的每条线段看成障碍物。现在来考虑一些特殊位置。
<p>&nbsp;<img alt="" src="http://www.cppblog.com/images/cppblog_com/wangzhihao/3329_0.JPG" height="400" width="600" border="0"></p>
<p>出题者应该会避免相切的数据。和当前木棍在同一直线上的特殊点，一种可以根据线段的中心确定它会不会阻挡。</p>
<p>&nbsp;</p>
<p>&nbsp;另外一种连中心也在同一条直线上。这时认为在有木棍实体的那侧的特殊点不会阻挡，而另一侧会。</p>
<p>&nbsp;</p>
<p>&nbsp;<img alt="" src="http://www.cppblog.com/images/cppblog_com/wangzhihao/3329_1.JPG" height="400" width="600" border="0"></p>
<p>总的来说，障碍物（一条线段）他可能出现的位置是<span>0~360</span>。而难区分的是<span>0</span>度和<span>360</span>度。如果发现一个点落在<span>0</span>度或<span>360</span>度的位置时，通过判断障碍物在左边（包括平行的时候）还是在右边来确定会不会阻挡。<br><br>还可以通过让木棍转动一个小的角度，看一下是远离障碍物，还是和障碍物相交来确定是否会阻挡。</p><img src ="http://www.cppblog.com/Wangzhihao/aggbug/110187.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-21 00:36 <a href="http://www.cppblog.com/Wangzhihao/articles/110187.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3525 Most Distant Point from the Sea</title><link>http://www.cppblog.com/Wangzhihao/articles/109785.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Mon, 15 Mar 2010 15:12:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/109785.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/109785.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/109785.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/109785.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/109785.html</trackback:ping><description><![CDATA[这道题用二分 + 半平面交去做可以. 但是不利于推广道凹多边形. <br>还有一种做法就是内切圆必然和多边形的某些边相切, 我们二分半径, 然后枚举两条相切的边(凹多边形也可能是顶点), 确定圆心, 再看多边形是否包含这个圆.<br>确定圆心的时候直接解方程就可以了.<br><br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/109785.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-15 23:12 <a href="http://www.cppblog.com/Wangzhihao/articles/109785.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3608 Bridge Across Islands</title><link>http://www.cppblog.com/Wangzhihao/articles/109777.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Mon, 15 Mar 2010 14:56:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/109777.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/109777.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/109777.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/109777.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/109777.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 旋转卡壳,&nbsp; 这道题的精度很奇怪,不知为什么转了一圈还不够. 我用叉积判断最小前进角 , 照理应该足够精确了. <br>当遇到这种精度卡的很紧的题目时,要尽量少用角度.<br><br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/109777.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-15 22:56 <a href="http://www.cppblog.com/Wangzhihao/articles/109777.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3675 Telescope</title><link>http://www.cppblog.com/Wangzhihao/articles/109773.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Mon, 15 Mar 2010 14:48:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/109773.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/109773.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/109773.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/109773.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/109773.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp; 求多边形和圆的面积交. 可转化为求三角形与圆的交这一小问题.这里的三角形的一个顶点是圆心.有四种情况.<br>对于每种情况.实际上只要算出圆和线段的相交区域的端点即可.<br><br><img alt="" src="http://www.cppblog.com/images/cppblog_com/wangzhihao/3675.jpg" height="400" width="640"><br><br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/109773.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-15 22:48 <a href="http://www.cppblog.com/Wangzhihao/articles/109773.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>2932 Coneology</title><link>http://www.cppblog.com/Wangzhihao/articles/109348.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Wed, 10 Mar 2010 05:05:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/109348.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/109348.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/109348.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/109348.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/109348.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: sweepline&nbsp;&nbsp;<a href='http://www.cppblog.com/Wangzhihao/articles/109348.html'>阅读全文</a><img src ="http://www.cppblog.com/Wangzhihao/aggbug/109348.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-10 13:05 <a href="http://www.cppblog.com/Wangzhihao/articles/109348.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3410 Split convex polygon</title><link>http://www.cppblog.com/Wangzhihao/articles/108911.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Thu, 04 Mar 2010 15:39:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/108911.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/108911.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/108911.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/108911.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/108911.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp; 这道题很显然不用求凸包, 应为给的两个多边形本身就具有很好的"序".<br>&nbsp;&nbsp;&nbsp; 首先判断两个多边形的凹凸性, 如果两个多边形都是凸的.那么两个凸多边形只会公共一条边,需要检查两端的凹凸性.<img style="width: 558px; height: 358px;" alt="" src="http://www.cppblog.com/images/cppblog_com/wangzhihao/3410_0.jpg"><br>&nbsp;如果有凹多边形,那么凹点必定是要和别人耦合的,可以先找一个凹点,再枚举另一个多边形的所有点,看是否和该凹点匹配,如果匹配,就沿着多边形的方向走,继续检查下一对点, 下一对点要么耦合, 要么以一个凸的形状分开.<br><br><img alt="" src="http://www.cppblog.com/images/cppblog_com/wangzhihao/3410_1.jpg" height="600" width="800"><br>最后只要检查所有的凹点是否都访问了就可以了.<br> <img src ="http://www.cppblog.com/Wangzhihao/aggbug/108911.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-04 23:39 <a href="http://www.cppblog.com/Wangzhihao/articles/108911.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3384 Feng Shui</title><link>http://www.cppblog.com/Wangzhihao/articles/108797.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Wed, 03 Mar 2010 07:44:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/108797.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/108797.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/108797.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/108797.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/108797.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp; 题目大意是给你一个凸多边形，然后用两个圆面去尽可能多的覆盖它，圆面必须包含在凸多边形内。两个圆面可以相交。求一对合法的圆心。<br>&nbsp;&nbsp; &nbsp;就是求凸多边形的核，再求核上的最远点。
<br> <img src ="http://www.cppblog.com/Wangzhihao/aggbug/108797.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-03 15:44 <a href="http://www.cppblog.com/Wangzhihao/articles/108797.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3347 Kadj Squares</title><link>http://www.cppblog.com/Wangzhihao/articles/108795.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Wed, 03 Mar 2010 07:29:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/108795.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/108795.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/108795.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/108795.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/108795.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp; 题目大意是有若干个大小不一（整数）的正方形，从左到右呈45`放置。相互紧靠，问从正上方能看到的正方形有哪些。最多50个正方形。<br>&nbsp;&nbsp; &nbsp;如果能确定每个正方形的位置。那么就可以很轻松的算出遮挡关系。这可以转换成一些区间的覆盖问题。<br>&nbsp;&nbsp; &nbsp;如果确定了1,2,...，k-1的位置，现在要确定第 k 个的位置。假设有两个正方形a，b。a的位置已经确定为 Xa，b在a的右边，那么 Xb = Xa + min(a, b)*sqrt(2);这样就可以确定第k块正方形的位置了。<br>&nbsp;&nbsp; &nbsp;注意到上面涉及浮点数，我们把边长扩大根号二倍，不影响最后结果，但只有整数间的运算。
<br>  <img src ="http://www.cppblog.com/Wangzhihao/aggbug/108795.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-03 15:29 <a href="http://www.cppblog.com/Wangzhihao/articles/108795.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3336 ACM Underground</title><link>http://www.cppblog.com/Wangzhihao/articles/108789.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Wed, 03 Mar 2010 07:00:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/108789.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/108789.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/108789.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/108789.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/108789.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 很训练代码能力&nbsp;&nbsp;<a href='http://www.cppblog.com/Wangzhihao/articles/108789.html'>阅读全文</a><img src ="http://www.cppblog.com/Wangzhihao/aggbug/108789.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-03-03 15:00 <a href="http://www.cppblog.com/Wangzhihao/articles/108789.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>3521 Geometric Map</title><link>http://www.cppblog.com/Wangzhihao/articles/108568.html</link><dc:creator>王之昊</dc:creator><author>王之昊</author><pubDate>Sat, 27 Feb 2010 11:03:00 GMT</pubDate><guid>http://www.cppblog.com/Wangzhihao/articles/108568.html</guid><wfw:comment>http://www.cppblog.com/Wangzhihao/comments/108568.html</wfw:comment><comments>http://www.cppblog.com/Wangzhihao/articles/108568.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/Wangzhihao/comments/commentRss/108568.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/Wangzhihao/services/trackbacks/108568.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 3521 Geometric Map&nbsp;&nbsp;<a href='http://www.cppblog.com/Wangzhihao/articles/108568.html'>阅读全文</a><img src ="http://www.cppblog.com/Wangzhihao/aggbug/108568.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/Wangzhihao/" target="_blank">王之昊</a> 2010-02-27 19:03 <a href="http://www.cppblog.com/Wangzhihao/articles/108568.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>