动态规划：PAT 1045 Favorite Color Stripe

ADS的第7周作业，是一个动态规划算法题。

问题重现

Title: Favorite Color Stripe

Description:

Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

It is said that a normal human eye can distinguish about less than 200 different colors, so Eva’s favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva’s favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N ( N≤200 ) which is the total number of colors involved (and hence the colors are numbered from 1 to N ). Then the next line starts with a positive integer M ( M≤200 ) followed by MM Eva’s favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L ( L≤104 ​​ ) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line a separated by a space.

Output Specification:

For each test case, simply print in a line the maximum length of Eva’s favorite stripe.

Sample Input:

6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7

解法说明

设Eva喜欢的序列为 M[0...m) ，给出条纹的序列为 L[0...l) 。
经过裁减，符合条件序列的最大长度（最大公共序列长度）的解为 f(m,l) 。
对于状态空间 {(m,l)|m∈M,l∈L} ，
平凡态：

f(0,l)=f(m,0)=0

有状态转移方程：

f(m,l)={f(m,l−1)+1,max{f(m−1,l),f(m,l−1)},Mm−1=Ll−1Mm−1≠Ll−1

平凡态比价容易理解：空序列的长度为0。

现在考虑当Eva喜欢的序列不变时，每当给出的条纹多了一个，即 Ll−1 ，考虑它是否与Eva喜欢的序列的最后一个( Mm−1 )相同，即它是否可以拓展，如果可以，它可以直接加到 f(m,l−1) 上，这个情况也比较好理解。

最后考虑当这个新来的条纹不匹配时，它，或者Eva喜欢的最后一个颜色，两者之一总有一个没有卵用的：当新来的条纹不存在( f(m,l−1) )或者Eva喜欢的最后一个颜色不存在( f(m−1,l) )，比较一下哪个比较大就取哪个即可。

代码实现

• 递归记忆化搜索版本
• 空间优化并使用迭代的版本
• 半在线算法进一步优化空间版本