<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[Notdeep__acm的专栏]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/Notdeep__acm</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; Notdeep__acm]]></copyright><item><title><![CDATA[POJ 2448(K短路,A*+SPFA) Remmarguts' Date]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47259949</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47259949</guid><author>Notdeep__acm</author><pubDate>Mon, 03 Aug 2015 18:05:33 +0800</pubDate><description><![CDATA[题意给一个n个点m条边的图,然后给一个起点和一个终点,求起点到终点的第K短路.思路求第K短路。一个经典的问题。SPFA+A* 
核心思想在A*搜索的估计函数的建立上。 
F(x) = g(x) + h(x) 
估价函数 = s到x的距离 + x到t的距离 
估价函数的含义就是经过x这个点的路径的距离。 
我们在搜索的时候每次选择估价函数较小的值,进行拓展。这样我们搜索到t点的状态出来顺序就是,最短路]]></description><category></category></item><item><title><![CDATA[hdu 5305(爆搜+二染色) Friends]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47157759</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47157759</guid><author>Notdeep__acm</author><pubDate>Thu, 30 Jul 2015 22:13:28 +0800</pubDate><description><![CDATA[题意：有n个人m对朋友,朋友有两种一种online的,一种face to face的。现在要让你对这些朋友性质进行安排,让每一个人的线上朋友和线下朋友人数是相同的,求满足的方案数。思路直接暴力染色。 
然后减减枝。 
一个就是对于每个人的度一定是偶数。 
另外一个就是染色中,对于一个节点染色的不能超过一半才能继续dfs。参考code:/*
 #pragma warning (disable: 478]]></description><category></category></item><item><title><![CDATA[hdu 5299(树上删边游戏) Circles Game]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47132011</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47132011</guid><author>Notdeep__acm</author><pubDate>Wed, 29 Jul 2015 18:22:58 +0800</pubDate><description><![CDATA[题意：给你一些圆圈的圆心坐标和半径,保证这些圆圈是包含或者相离的。现在两个人做博弈操作,拿掉一个那么这个所包含的都要移除掉。两个人一直拿下去,直到某个人找不到一个可以移除的圈他就输了。思路学习的姿势。这是一个树上删边游戏模型； 
首先按照半径升序排序,然后对于每一个圆找第一个包含它的圆,然后连一条边。建树完成。 
叶子节点的SG值为0; 
中间节点的 SG 值为它的所有子节点的 SG 值加 1 后的]]></description><category></category></item><item><title><![CDATA[hdu 5289(二分+RMQ) Assignment]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47113443</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47113443</guid><author>Notdeep__acm</author><pubDate>Tue, 28 Jul 2015 22:42:31 +0800</pubDate><description><![CDATA[题意：给一个序列,然后求出连续的序列中最大和最小值之差小于k的。思路二分+ST 
二分下标,然后找一个最大的区间满足区间内最大最小值相差小于k,当前这个位置对于答案的贡献就是这个区间长度。 
求一个静态数组的区间最大最小值,用ST算法就好了。参考code:/*
 #pragma warning (disable: 4786)
 #pragma comment (linker, "/STACK:0x8]]></description><category></category></item><item><title><![CDATA[hdu 5288 (预处理+暴力) OO’s Sequence]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47112769</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/47112769</guid><author>Notdeep__acm</author><pubDate>Tue, 28 Jul 2015 22:09:36 +0800</pubDate><description><![CDATA[题意：见题面。思路预处理出每个a[i]左边和右边第一个能整除它的位置L[i]和R[i],然后计算这个值对于答案的贡献的个数。 
贡献就是左右区间长度相乘参考code:/*
 #pragma warning (disable: 4786)
 #pragma comment (linker, "/STACK:0x800000")
 */
#include <cassert>
#include <ccty]]></description><category></category></item><item><title><![CDATA[hdu 3466(排序+背包)Proud Merchants]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46903087</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46903087</guid><author>Notdeep__acm</author><pubDate>Thu, 16 Jul 2015 01:17:26 +0800</pubDate><description><![CDATA[题意：给n个物品,每个物品有三个属性,价格、你至少拥有钱数才能购买该物品的值和购买后能获得的价值。然后你有m块钱,问最多能获得多少价值。思路看起来就是一个01背包加了一个购买钱数的限制。 
就用01肯定不行。 
用极限来看裸的01背包的限制就是它的的价格,相当于限制值为0,现在给了一个新限制,那么我对这个限制值进行一个从小到大的排序,然后01背包做过去就行了。这样前面就不会影响后面的状态了。 
很好]]></description><category></category></item><item><title><![CDATA[hdu 3449(依赖背包) Consumer]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46902325</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46902325</guid><author>Notdeep__acm</author><pubDate>Thu, 16 Jul 2015 00:44:08 +0800</pubDate><description><![CDATA[题意：给n个箱子,每个箱子有个费用pi,每个箱子里面有m个物品,每个物品有两个属性,一个费用wi和一个价值vi。买物品的时候要把所属的箱子一起买了,问给一个金额W最多能获得多少价值?思路依赖背包。《背包九讲》的第七讲讲了这个。 
设dp[i][j]为前i个箱子使用金额j获得的最大价值。 
考虑每个箱子内部就是一个01背包,那么在枚举每一个箱子的时候,使用上一轮能满足本轮箱子费用的状态做一个01背包,]]></description><category></category></item><item><title><![CDATA[hdu 3448(搜索+剪枝)Bag Problem]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46901061</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46901061</guid><author>Notdeep__acm</author><pubDate>Wed, 15 Jul 2015 23:47:38 +0800</pubDate><description><![CDATA[题意：给n(n<=40)个物品和背包的容量w以及背包能装的物品个数k,每个物品有一个重量,问在满足背包的限制的情况下最多可以装多少物品。思路做过类似的题目,第一反应就是爆搜每个物品的两个状态放和不放。2^40肯定不行，来剪枝吧。 
先把物品从小到大排序。一个有效的剪枝就是,最大的k个物品的重量和小于w那么这个重量和就是答案了,也是搜索中比较极限的情况,避免了去做搜索。 
这样可以水过了,几乎没跑时间]]></description><category></category></item><item><title><![CDATA[hdu 3433(dp+二分) A Task Process]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46898723</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/46898723</guid><author>Notdeep__acm</author><pubDate>Wed, 15 Jul 2015 21:11:54 +0800</pubDate><description><![CDATA[题意：有n个人，每个人完成一件A任务需要a[i]时间，完成一件B任务需要b[i]时间，问要完成x件A任务y件B任务所需的最短时间是多少?思路二分+DP。 
二分时间t,dp[i][j]表示在时间t内前i个人完成j件A任务所能完成的B任务的最大数量。如果dp[i][x]>=y 
就是可以的。然后不断迭代得到ans。复杂度:0(n*x^2*log(t)) t范围是40w参考code:/*
 #pragm]]></description><category></category></item><item><title><![CDATA[Codeforces Round #280 (Div. 2) D]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669865</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669865</guid><author>Notdeep__acm</author><pubDate>Tue, 02 Dec 2014 03:40:45 +0800</pubDate><description><![CDATA[Codeforces Round #280 (Div. 2) D]]></description><category></category></item><item><title><![CDATA[Codeforces Round #280 (Div. 2) C]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669701</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669701</guid><author>Notdeep__acm</author><pubDate>Tue, 02 Dec 2014 03:26:22 +0800</pubDate><description><![CDATA[Codeforces Round #280 (Div. 2) C]]></description><category></category></item><item><title><![CDATA[Codeforces Round #280 (Div. 2) B]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669587</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669587</guid><author>Notdeep__acm</author><pubDate>Tue, 02 Dec 2014 03:22:46 +0800</pubDate><description><![CDATA[Codeforces Round #280 (Div. 2) B]]></description><category></category></item><item><title><![CDATA[Codeforces Round #280 (Div. 2) A]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669481</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/41669481</guid><author>Notdeep__acm</author><pubDate>Tue, 02 Dec 2014 03:21:15 +0800</pubDate><description><![CDATA[Codeforces Round #280 (Div. 2) A]]></description><category></category></item><item><title><![CDATA[Codeforces Round #275 (Div. 2) C]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445831</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445831</guid><author>Notdeep__acm</author><pubDate>Sat, 25 Oct 2014 03:49:34 +0800</pubDate><description><![CDATA[Codeforces Round #275 (Div. 2) C]]></description><category></category></item><item><title><![CDATA[Codeforces Round #275 (Div. 2) B]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445731</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445731</guid><author>Notdeep__acm</author><pubDate>Sat, 25 Oct 2014 03:43:36 +0800</pubDate><description><![CDATA[Codeforces Round #275 (Div. 2) B]]></description><category></category></item><item><title><![CDATA[Codeforces Round #275 (Div. 2) A]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445665</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40445665</guid><author>Notdeep__acm</author><pubDate>Sat, 25 Oct 2014 03:40:38 +0800</pubDate><description><![CDATA[Codeforces Round #275 (Div. 2) A]]></description><category></category></item><item><title><![CDATA[Codeforces Round #272 (Div. 2) C]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051693</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051693</guid><author>Notdeep__acm</author><pubDate>Mon, 13 Oct 2014 20:35:32 +0800</pubDate><description><![CDATA[Codeforces Round #272 (Div. 2) C]]></description><category></category></item><item><title><![CDATA[Codeforces Round #272 (Div. 2) B]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051625</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051625</guid><author>Notdeep__acm</author><pubDate>Mon, 13 Oct 2014 20:32:04 +0800</pubDate><description><![CDATA[Codeforces Round #272 (Div. 2) B]]></description><category></category></item><item><title><![CDATA[Codeforces Round #272 (Div. 2) A]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051543</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/40051543</guid><author>Notdeep__acm</author><pubDate>Mon, 13 Oct 2014 20:25:53 +0800</pubDate><description><![CDATA[Codeforces Round #272 (Div. 2) A]]></description><category></category></item><item><title><![CDATA[Codeforces Round #267 (Div. 2) A]]></title><link>https://blog.youkuaiyun.com/Notdeep__acm/article/details/39388303</link><guid>https://blog.youkuaiyun.com/Notdeep__acm/article/details/39388303</guid><author>Notdeep__acm</author><pubDate>Fri, 19 Sep 2014 02:29:47 +0800</pubDate><description><![CDATA[Codeforces Round #267 (Div. 2) A]]></description><category></category></item></channel></rss>