上题目……本来以为这道题目不用写个日志了的……最后还是悲剧……动规做得少了最近……竟然范这种低级错误……

Gangsters

Description

N gangsters are going to a restaurant. The i-th gangster comes at the time Ti and has the prosperity Pi. The door of the restaurant has K+1 states of openness expressed by the integers in the range [0, K]. The state of openness can change by one in one unit of time; i.e. it either opens by one, closes by one or remains the same. At the initial moment of time the door is closed (state 0). The i-th gangster enters the restaurant only if the door is opened specially for him, i.e. when the state of openness coincides with his stoutness Si. If at the moment of time when the gangster comes to the restaurant the state of openness is not equal to his stoutness, then the gangster goes away and never returns.

The restaurant works in the interval of time [0, T].

The goal is to gather the gangsters with the maximal total prosperity in the restaurant by opening and closing the door appropriately.

Input

?The first line of the input file contains the values N, K, and T, separated by spaces. (1 <= N <= 100 ,1 <= K <= 100 ,0 <= T <= 30000 )

?The second line of the input file contains the moments of time when gangsters come to the restaurant T1, T2, …, TN, separated by spaces. ( 0 <= Ti <= T for i = 1, 2, …, N)

?The third line of the input file contains the values of the prosperity of gangsters P1, P2, …, PN, separated by spaces. ( 0 <= Pi <= 300 for i = 1, 2, …, N)

?The forth line of the input file contains the values of the stoutness of gangsters S1, S2, …, SN, separated by spaces. ( 1 <= Si <= K for i = 1, 2, …, N)

All values in the input file are integers.

Output

Print to the output file the single integer ?the maximal sum of prosperity of gangsters in the restaurant. In case when no gangster can enter the restaurant the output should be 0.

Sample Input

4 10 20<br></br>10 16 8 16<br></br>10 11 15 1<br></br>10 7 1 8<br></br>

Sample Output

26<br></br><br></br>题目意思有点难懂……感觉……看了好几次才看懂到底啥意思，英语水平不够还……其实意思就是门的值没一个time unit可以＋1，－1或者不变，然后对于某一个客人，必须在门的值跟他自己所拥有的一个值相等时才进门，要求总体的P相加最大……典型的动规……当即就想到了方程，直接写程序……写了20分钟吧……感觉最近用VIM写了这么多个程序，速度快了些了……<br></br>然后提交，发现尽然MLE了……仔细一看，发现数组是开大了……于是改成循环数组，循环数组已经很熟了反正……改好再提交，尽然WA了……看了好几遍代码……没有发现错误……最后看时间不早了，就去了Discuss……把里面的测试数据都试了一遍，还是没有找到错误……然后发现这道题目跟上次下的官方数据是一个系列的……于是就拿官方数据测试……发现问题所在了……<br></br>问题在于初始化的时候没有把k＋1的位置初始化，然后在动规的时候又没有检查边界……<br></br>这个问题其实我有两次机会可以避免……第一是初始化的时候……不知道咋想的没有初始化完……话说数组都特意开大了的……以后初始化一定多初始化几位……<br></br>第二是在动规的时候其实可以检查一下边界……对于下界我特意注意过了……对于上界竟然忽略了……不知道当时写代码的时候怎么想的……其实考虑过可能会出问题……但是就是没有去动它……想着少写几行代码……<br></br>不过话说回来……一开始那种单独写一个循环特殊处理下界的办法其实挺dirty的……虽然速度会快一点……dirty的代价就是没有考虑周全……哎……<br></br>话说这么简单的题目……悲剧……多做做题目应该会好些<br></br>