reward

扩展KMP --- HDU 3613 Best Reward

 Best Reward Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?
·2015-10-28 08:32

HDU Best Reward (扩展KMP)

Best Reward Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/65536K (Java/
·2015-10-27 16:12

POJ 3621 Sightseeing Cows （参数搜索 + 最短路(ms 原始的bellman tle, 用spfa才过)

Total Submissions: 6684   Accepted: 2196 Description Farmer John has decided to reward
·2015-10-23 08:27

HDU-2647 Reward ---拓扑排序

Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java
·2015-10-23 08:32

HDU 2647 Reward

描述 Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards. The workers
·2015-10-23 08:54

hdu 3613 exkmp

pid=3613 Best Reward Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536
·2015-10-21 11:18

Andrew Ng机器学习课程17(2)

blog.csdn.net/lg1259156776/说明：主要介绍了利用valueiteration和policyiteration两种迭代算法求解MDP问题，还介绍了在实际应用中如何通过积累“经验”更新对转移概率和reward
LG1259156776·2015-10-11 21:00

HDOJ 2647 Reward(反拓扑+统计节点)

题目大意：有一种关系a,b代表a的奖金比b的高，最少的奖金为888，求总共需要的最少的钱数，让所有满足关系的人都得到应有的奖金，否则输出-1；思路：反向进行拓扑，主要是统计奖金。注意可能给定的数据是深林，所以可以先统计出度为0的点赋值为888，然后边拓扑边统计。#include #include #include #include #include #definell__int64 usingna
Grit_ICPC·2015-10-03 23:00

表链接后进行计算

selectt.id,t.userId,taskType,reward,servicesTime,timeLength,publicTime,notes,location,headPicture,tu.counts
yangxujia·2015-09-18 15:00

Reward for virtue

原文Myfriend,Hugh,hasalwaysbeenfat,butthingsgotsobadrecentlythathedecidedtogoonadiet.Hebeganhisdietaweekago.Firstofall,hewroteoutalonglistofallthefoodswhichwereforbidden.ThelistincludedmostofthethingsHu
Robinson_lu·2015-09-18 12:00

hdu 2647 Reward

clickhere~~Reward TimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others) TotalSubmission
qingshui23·2015-08-17 19:00

Reward 2647 （拓扑排序+邻接表）

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):5701    AcceptedSubmission(s):1730ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
yanghui07216·2015-08-14 21:00

HDU-2647 Reward（拓扑排序）

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfestivaliscoming,hewantstodistributerewardstohisworkers.
idealism_xxm·2015-08-14 20:00

HDOJ 2647 Reward（拓扑排序（节点含值）+邻接表）

RewardTimeLimit:2000/1000MS(Java/Others)MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):5678AcceptedSubmission(s):1720ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfestivalisc
AnICoo1·2015-08-14 17:40

HDOJ 2647 Reward（拓扑排序（节点含值）+邻接表）

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):5678    AcceptedSubmission(s):1720ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
helloiamclh·2015-08-14 17:00

HDU 2647--Reward【拓扑排序】

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):5636    AcceptedSubmission(s):1712ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
hpuhjh·2015-08-14 12:00

HDU 3613 Best Reward（manacher)

BestRewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/65536K(Java/Others)TotalSubmission(s):1074    AcceptedSubmission(s):438ProblemDescriptionAfteranuphillbattle,GeneralLiwonagreatvictory
dml_96·2015-08-14 10:00

HDU2647 Reward 拓扑排序

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2647题目大意：有n个人m对关系，每对关系表示a的工资要比b的高，最低工资为888，问你这n个人满足m对关系的最低总工资是多少。分析：我们以这m对关系来建图，要使a的工资要比b的工资高，只需保证a到b的单向性即可。实现代码如下：#include #include #include #include us
AC_Gibson·2015-08-11 10:00

Hdoj 2467 Reward(拓扑排序)

DescriptionDandelion'suncleisabossofafactory.Asthespringfestivaliscoming,hewantstodistributerewardstohisworkers.Nowhehasatroubleabouthowtodistributetherewards. Theworkerswillcomparetheirrewards,andsom
just_sort·2015-08-10 16:00

HDU Reward(发工资) 2647 （拓扑排序 vector）

ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfestivaliscoming,hewantstodistributerewardstohisworkers.Nowhehasatroubleabouthowtodistributetherewards.Theworkerswillcomparetheirrewards,
h1021456873·2015-08-03 16:00

hdu 2647 Reward 拓扑排序

#include #include #include #include #include #include usingnamespacestd; vectort[10000+5]; queueq; intmain() { intn,m,i,j,in[10000+5],sum[10000+5],ans,u,v,p,cnt; while(~scanf("%d%d",&n,&m)) { ans=n*8
xinag578·2015-07-30 10:00

hdoj 2647 Reward 【反向拓扑 队列实现】

RewardTimeLimit:2000/1000MS(Java/Others)   MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4767   AcceptedSubmission(s):1456ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfesti
chenzhenyu123456·2015-05-21 13:00

HDU_2647_Reward(拓扑排序)

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4746    AcceptedSubmission(s):1448ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
jhgkjhg_ugtdk77·2015-05-07 20:00

HDU 2647 Reward

ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfestivaliscoming,hewantstodistributerewardstohisworkers.Nowhehasatroubleabouthowtodistributetherewards.Theworkerswillcomparetheirrewards,
jtjy568805874·2015-05-03 09:00

hdu 2647 Reward 拓扑排序。

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4599    AcceptedSubmission(s):1400ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
Lionel_D·2015-03-28 10:00

HDU 3613 Best Reward(求前后缀回文 拓展KMP or Manacher)

题目大意：给个字符串X，要把X分成两段T1,T2，每个字母都有一个对应的价值，如果T1，T2是回文串（从左往右或者从右往左读，都一样），那么他们就会有一个价值，这个价值是这个串的所有字母价值之和，如果不是回文串，那么这串价值就为0。问最多能获得多少价值？思路：把字符串X逆序后得到字符串Y让X去匹配Y，匹配的长度满足extend[i]+i==len, len=|X|.  的那么X与y的匹配部分是回文
kalilili·2015-02-21 15:00

hdu 3613 Best Reward (kmp扩展)

题意：给出一个字符串，并且给出每种字符串对应的价值，现在要二分这个字符串，如果二分的两部分，如果是回文串就加上对应字符的价值所谓这两个部分的价值，如果不是回文两个部分的价值为0。题解：两种方法：法一：这题相当于可以转化为求回文前缀和后缀,求回文前缀：将原串反转接到原串后面。求回文后缀：将原串接到原串反转后的串后面。人后求next数组，next[len]就最长的那个回文前缀，不断k=next[k]就
My_ACM_Dream·2015-02-12 22:00

hdu3613---Best Reward

BestRewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/65536K(Java/Others)TotalSubmission(s):771    AcceptedSubmission(s):333ProblemDescriptionAfteranuphillbattle,GeneralLiwonagreatvictory.
Guard_Mine·2015-02-02 12:00

HDU2647 Reward 【拓扑排序】

RewardTimeLimit:2000/1000MS(Java/Others)  MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4379  AcceptedSubmission(s):1335ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfestiva
u011676797·2014-12-22 15:00

HDU 2647 Reward 【逆向建图+拓扑排序】

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4337    AcceptedSubmission(s):1328ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
u013806814·2014-11-26 19:00

httpinvoker异常

CouldnotdeserializeresultfromHTTPinvokerremoteservice[http://localhost:9080/];nestedexceptionisjava.io.InvalidClassException:com.reward.domain.user
mrliuze·2014-11-20 17:00

hdu2647——Reward

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4306    AcceptedSubmission(s):1318ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
Guard_Mine·2014-10-29 16:00

Reward

Reward时间限制：1000 ms | 内存限制：65535 KB难度：3描述Dandelion'suncleisabossofafactory.Asthespringfestivaliscoming
u012629369·2014-10-20 21:00

magento rewrite enterprise model

myreward> <class>Bysoft_Myreward_Model</class> </myreward> <enterprise_reward
kirenenko04·2014-09-16 10:00

关于习惯

习惯形成有三个组成部分：cue、routine、reward。 形成习惯需要遵循一定规则，打破习惯也需要遵循一定规则（破坏掉某个组成部分）。
·2014-08-25 11:00

HDU2647 Reward 【拓扑排序】

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):3927    AcceptedSubmission(s):1199ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfes
u012846486·2014-08-02 08:00

hdu 3613 Best Reward回文子串Manocher算法

ProblemDescriptionAfteranuphillbattle,GeneralLiwonagreatvictory.Nowtheheadofstatedecidetorewardhimwithhonorandtreasuresforhisgreatexploit.Oneofthesetreasuresisanecklacemadeupof26differentkindsofgemsto
firenet1·2014-08-01 14:00

mysql问题解决

SELECT a.Pid,a.id FROM `user` as b,`reward` as a WHERE b.id=4 and a.Pid = b.Pid;在执行上面语句是报错：Illegalmixofcollations
风亭了云知道·2014-07-09 13:00

hdu Reward(拓扑排序)

N- RewardTimeLimit:1000MS     MemoryLimit:32768KB     64bitIOFormat:%I64d&%I64uSubmit StatusDescriptionDandelion'suncleisabossofafactory.Asthespringfestivaliscoming,hewantstodistributerewardstohiswork
u014569598·2014-07-08 19:00

【KMP】 HDOJ 3613 Best Reward

做了很久，做不出来，只好看题解了。。。要注意到分割后的两个串是有特点的。前一个串一定包含原始串的第一个字符，后一个串一定包含原始串最后一个字符。如果能注意到就可以应该能想出来了。。。把原始串记为S1，将原始串翻转记为S2。然后先将S1作为模式串，用S1匹配S2，最后得到的最大匹配就是S1串前缀最长回文串，利用next数组就可以求出所有S1串前缀回文串。。。同样的将S2作为模式串，用S2匹配S1，最
blankcqk·2014-06-29 15:00

HDU 2647 Reward(拓扑排序)

HDU2647Reward(拓扑排序)http://acm.hdu.edu.cn/showproblem.php?
u013480600·2014-06-14 11:00

cf枪王排位段位等级奖励道具领取 段位等级奖励大全

cf枪王排位段位等级奖励道具领取网址：http://cf.qq.com/rank/reward.shtml段位等级奖励领取说明：在本次测试期间（6月5日至6月25日），达到相应等级的玩家可直接领取以下道具职业佣兵奖励内容领取
佚名·2014-06-06 08:40

HDU 2647Reward(拓扑排序）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=2647去了趟厕所回来立马来灵感了啊。。第一次的时候纠结拓扑的顺序，加钱的时候顺序正好反着，在厕所的时候突然想到可以在加边的时候就倒着加，这样不就正好了吗。。于是回来怒敲代码，然后1Y，爽啊。。于是就多写点注释吧。。#include #include #include #include #include
u013013910·2014-06-01 15:00

HDU 3613 Best Reward(扩展KMP:回文判断)

HDU3613BestReward(扩展KMP:回文判断)http://acm.hdu.edu.cn/showproblem.php?pid=3613题意:给你一个字符串,要你把这个字符串分成两段,并使得被分开的两段价值和最大.一个串如果是回文,那么它的价值就是所有字符的价值和,否则价值为0.分析:      首先原始串为S,将S逆转得到串T.(S=abcaaa,那么T=aaacba).     
u013480600·2014-04-06 20:00

HDOJ 3613 Best Reward

本来想练EX_KMP的，一读题发现是回文串，与回文串有关的一定要用神器manacher啊！！！然后就用manacher写了。。manacher出所有的回文串半径，然后预处理一遍，最后统计最大值。从左边延续过来的回文串，要看右边有没有构成回文，反之亦然。。。BestRewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/65536K
u012797220·2014-04-01 17:00

award vs reward

Award:Anawardisanhonor(suchasamedalortrophy)thatisconferredformerit,usuallyafteradecisionmadebyacommitteeofjudgesReward:Arewardissomething(usuallymoney)giveninreturnforaperson'sservice,merit,hardship,
gentle_wolf·2014-02-20 12:00

unity3d xml序列化

using System.Xml.Serialization; using System.IO; using System.Text; public class Reward
·2013-10-15 22:00

hdu2647 Reward（拓扑排序）

老板要给很多员工发奖金，但是部分员工有个虚伪心态，认为自己的奖金必须比某些人高才心理平衡；但是老板很人道，想满足所有人的要求，并且很吝啬，想画的钱最少输入若干个关系abaccb意味着a的工资必须比b的工资高同时a的工资比c高；c的工资比b高当出现环的时候输出-1思路：反向建图，然后top排序分层次；第一次的工资为888（最低），第二层的工资+1，后面一样#include #include #inc
ygqwan·2013-09-19 09:00

HDU 2647 Reward

大体思路就是简单的拓扑排序。写的时候没有看模板，都是自己一点一点改的，应该没有模板效率高。还用了一点优先队列的思想，用了并查集的方法检查是否存在环。1#include 2#include 3#include 4 5usingnamespacestd; 6 7structN 8{ 9intdata; 10intn,money; 11N*next; 12}*head[10010]
u012161037·2013-09-18 15:00

hdu2647 && TOJ3648 Reward

RewardTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):2862    AcceptedSubmission(s):844ProblemDescriptionDandelion'suncleisabossofafactory.Asthespringfest
wangwenhao00·2013-08-27 23:00
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