﻿<?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-The Sun Also Rises-文章分类-ACM/ICPC</title><link>http://www.cppblog.com/FreePeter/category/6172.html</link><description>Algorithm, Mathematica, 计算机科学, C++, photography, Linux的讨论空间</description><language>zh-cn</language><lastBuildDate>Wed, 21 May 2008 09:01:25 GMT</lastBuildDate><pubDate>Wed, 21 May 2008 09:01:25 GMT</pubDate><ttl>60</ttl><item><title>[RESTROSPECT] Anton Golubev (Hedgehog)'s Contest #2 from Annual Summer Russian Teams Meeting in Petrozavodsk State University, SGU 287-295</title><link>http://www.cppblog.com/FreePeter/articles/Retrospect_To_SGU287_295.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Thu, 14 Feb 2008 05:02:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Retrospect_To_SGU287_295.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42725.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Retrospect_To_SGU287_295.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42725.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42725.html</trackback:ping><description><![CDATA[287&nbsp;&nbsp; &nbsp; Amusing Qc Machine<br>很有意思的一道题。<br>f[i] = i次猜测可以确定的范围。<br>首先我们猜一个，如果得到了right就over了是吧。<br>否则我们还不知道应该往哪边猜，先蒙一个方向猜，这样可以确定的范围是f[i - 1]<br>然后c次以后得到了正确的方向，如果是另一个方向那么能确定的范围是f[i - c]<br>so, f[i] = f[i - 1] + f[i - c] + 1...<br><br>288&nbsp;&nbsp; &nbsp; Best Tournament Schedule&nbsp;&nbsp;&nbsp; <br>经典的构造题<br><br>289&nbsp;&nbsp; &nbsp; Challenging Tic-Tac-Toe<br>经典的博弈树搜索 + memorize<br>&nbsp;&nbsp;&nbsp; <br>290&nbsp;&nbsp; &nbsp; Defend the Milky Way&nbsp;&nbsp;&nbsp; <br>三维凸包<br><br>291&nbsp;&nbsp; &nbsp; Evolution&nbsp;&nbsp;&nbsp; <br>用queue来模拟，最多只有1000*1000个繁殖事件。<br><br>292&nbsp;&nbsp; &nbsp; Field for the Cemetery <br><strong>q*c的格子最多可以放多少个1*c的长条(sgu292)</strong><br style="FONT-WEIGHT: bold">基本YY是只有2种可能，一种是使劲摆横然后摆竖，另一种是弦图那样摆<br>for e.g. 7 * 6放1*4<br><span style="FONT-FAMILY: Courier">aaaahi</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">bbbbhi</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">cccchi</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">de&nbsp; hi</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">degggg</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">dehhhh</span><br style="FONT-FAMILY: Courier"><span style="FONT-FAMILY: Courier">deiiii</span><br>伪代码:<br style="FONT-STYLE: italic"><span style="FONT-STYLE: italic">if (q &lt; n || c &lt; n) 直接算;</span><br style="FONT-STYLE: italic"><span style="FONT-STYLE: italic">t1 = q % n, t2 = c % n;</span><br style="FONT-STYLE: italic"><span style="FONT-STYLE: italic">s1 = n - t1, s2 = n - t2;</span><br style="FONT-STYLE: italic"><span style="FONT-STYLE: italic">ans = max((q * c - t1 * t2) / n, (q * c - s1 * s2) / n); </span><br style="FONT-STYLE: italic"><span style="FONT-STYLE: italic">&nbsp;&nbsp; </span><br>293&nbsp;&nbsp; &nbsp; Game with Q an C&nbsp;&nbsp;&nbsp; <br>应该是很麻烦的构造，还没想好。<br><br>294&nbsp;&nbsp; &nbsp; He's Circles&nbsp;&nbsp;&nbsp; <br>Polya原理<br><br>295&nbsp;&nbsp; &nbsp; Identifier Duplicated!<br>分别计算latin-russian和空格放置的方案数，乘起来。后者是不定方程的解数。<br>&nbsp;&nbsp;&nbsp; <br><br><br>
<img src ="http://www.cppblog.com/FreePeter/aggbug/42725.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-02-14 13:02 <a href="http://www.cppblog.com/FreePeter/articles/Retrospect_To_SGU287_295.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Northwestern Europe (NWERC) 2007 解题报告</title><link>http://www.cppblog.com/FreePeter/articles/Solution_To_NWERC2007.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Sat, 09 Feb 2008 11:10:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Solution_To_NWERC2007.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42606.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Solution_To_NWERC2007.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42606.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42606.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 这套题比较。。。有意思。。。Assemble据说是简单题March of the Penguins你看相传每个ice floes是有跳的次数限制的是吧。。。用经典的拆点流量限制法，每个ice floes拆成i和i'，i到i'的流量上限为跳的次数，然后如果能从i跳到j连一条i'到j的边(流量无穷大)。。。点数好像有点多。。。不要客气～，dinic很快的～Containers推一下式子，在实数意义下可...&nbsp;&nbsp;<a href='http://www.cppblog.com/FreePeter/articles/Solution_To_NWERC2007.html'>阅读全文</a><img src ="http://www.cppblog.com/FreePeter/aggbug/42606.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-02-09 19:10 <a href="http://www.cppblog.com/FreePeter/articles/Solution_To_NWERC2007.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Central Europe 2005解题报告</title><link>http://www.cppblog.com/FreePeter/articles/Solution_To_CERC2005.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Sun, 03 Feb 2008 13:32:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Solution_To_CERC2005.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42487.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Solution_To_CERC2005.html#Feedback</comments><slash:comments>2</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42487.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42487.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: AKnights of the Round Table模型：给出一张图(|V| &lt;= 1000)，求所有包含在奇数环中的点。算法分析：对每一个连通的子图，找出图中的割点，然后这些割点可以将图分为几部分，对于每一部分，每个点都至少在一个环中，因此如果该部分存在一个奇数环，则该部分所有点都属于一个奇数环。对于某一个图中是否存在奇数环的问题，只需黑白染色判断是否是二分图即可~复杂度|V| ...&nbsp;&nbsp;<a href='http://www.cppblog.com/FreePeter/articles/Solution_To_CERC2005.html'>阅读全文</a><img src ="http://www.cppblog.com/FreePeter/aggbug/42487.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-02-03 21:32 <a href="http://www.cppblog.com/FreePeter/articles/Solution_To_CERC2005.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Beijing 2002解题报告</title><link>http://www.cppblog.com/FreePeter/articles/Solution_To_Beijing_2002.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Wed, 30 Jan 2008 19:11:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Solution_To_Beijing_2002.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42262.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Solution_To_Beijing_2002.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42262.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42262.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 注：本文根据以前笔记整理而成，若有问题可以留言 or 询问我～Finding NemoBFS，我用了SPFASearching the Web模拟题，我用了一堆STL~~~Argus用个堆来维护一下就行了。Fun Game做的真辛苦@@@厄，首先如果只有1个方向并且是形成链的话很容易想出O(2 ^ 16 * 16 * 16)的算法2个方向的问题容易解决，同时考虑正向和反向，复杂度变...&nbsp;&nbsp;<a href='http://www.cppblog.com/FreePeter/articles/Solution_To_Beijing_2002.html'>阅读全文</a><img src ="http://www.cppblog.com/FreePeter/aggbug/42262.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-01-31 03:11 <a href="http://www.cppblog.com/FreePeter/articles/Solution_To_Beijing_2002.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Central Europe 2007解题报告</title><link>http://www.cppblog.com/FreePeter/articles/Solution_To_Central_Europe_2007.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Mon, 28 Jan 2008 07:49:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Solution_To_Central_Europe_2007.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42004.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Solution_To_Central_Europe_2007.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42004.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42004.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 比较不错的一套题，很多题挺有意思的～Strange Billboard经典思路了，枚举第一行的使用方法（2^16），然后推后面的方案。Cell Phone以每个点为圆心，r为半径画圆，问题转化为求被覆盖次数最多的区域次数。可以在每个圆上将相交的圆弧求出来，排序扫描，复杂度O(n^2logn)CEOI06 Antenna有一种解法就是二分半径然后用这个方法来求最多能覆盖多少个点。CODECo...&nbsp;&nbsp;<a href='http://www.cppblog.com/FreePeter/articles/Solution_To_Central_Europe_2007.html'>阅读全文</a><img src ="http://www.cppblog.com/FreePeter/aggbug/42004.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-01-28 15:49 <a href="http://www.cppblog.com/FreePeter/articles/Solution_To_Central_Europe_2007.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item><item><title>Southeastern Europe 2007解题报告</title><link>http://www.cppblog.com/FreePeter/articles/Solution_To_Southeastern_Europe_2007.html</link><dc:creator>FreePeter</dc:creator><author>FreePeter</author><pubDate>Sun, 27 Jan 2008 16:43:00 GMT</pubDate><guid>http://www.cppblog.com/FreePeter/articles/Solution_To_Southeastern_Europe_2007.html</guid><wfw:comment>http://www.cppblog.com/FreePeter/comments/42002.html</wfw:comment><comments>http://www.cppblog.com/FreePeter/articles/Solution_To_Southeastern_Europe_2007.html#Feedback</comments><slash:comments>0</slash:comments><wfw:commentRss>http://www.cppblog.com/FreePeter/comments/commentRss/42002.html</wfw:commentRss><trackback:ping>http://www.cppblog.com/FreePeter/services/trackbacks/42002.html</trackback:ping><description><![CDATA[&nbsp;&nbsp;&nbsp;&nbsp; 摘要: 经典题/模型比较多John&nbsp;&nbsp;&nbsp;Recommend经典的Mis`ere Nim问题，参见Impartial Combinatorial Games.2.5。非常精巧的一个思路。对于一个至少有2堆都&gt;1的且Nim和非0的状态，我们按照普通Nim的策略来进行。因为至少有2堆&gt;1，所以一步之内我们仍然至少有1堆&gt;1，又注意到我们留给对手的是一个平衡态，...&nbsp;&nbsp;<a href='http://www.cppblog.com/FreePeter/articles/Solution_To_Southeastern_Europe_2007.html'>阅读全文</a><img src ="http://www.cppblog.com/FreePeter/aggbug/42002.html" width = "1" height = "1" /><br><br><div align=right><a style="text-decoration:none;" href="http://www.cppblog.com/FreePeter/" target="_blank">FreePeter</a> 2008-01-28 00:43 <a href="http://www.cppblog.com/FreePeter/articles/Solution_To_Southeastern_Europe_2007.html#Feedback" target="_blank" style="text-decoration:none;">发表评论</a></div>]]></description></item></channel></rss>