PAT甲级：1045 Favorite Color Stripe

题目描述：

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 (≤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 (≤200) followed by M Eva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integer 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

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

题目大意： 

给定a, b字符串，求出a b两个字符串公共子序列的最大长度。其中a子序列中的某个数字可以按照自身的顺序在公共子序列中出现多次。

解题思路： 

动态规划

状态 dp[i][j] 的集合位对于a前i个数字和b前j个数字所有的公共子序列，其值代表这些子序列的长度最大值。
状态转移方式和最长公共子序列问题类似，考虑a的第i个数字以及b的第j个数字是否在子序列中出现分为四种情况进行讨论。

不同的是，对于第i个数字和第j个数字同时出现在子序列的情况，由于a中的数字可以在公子序列中出现多次，当满足b[j] == a[i]时，状态dp[i][j]由dp[i][j - 1]转移而来。

Python3代码： 

import sys
N = int(input())
lst1 = list(map(int,sys.stdin.readline().split()))
M = lst1[0] ; lst1 = [0] + lst1[1:]
lst2 = list(map(int,sys.stdin.readline().split()))
L = lst2[0] ; lst2 = [0] + lst2[1:]
 
dp = [[0]*(L+10) for i in range(M+10)]
for i in range(1,M+1) :
    for j in range(1,L+1) :
        dp[i][j] = dp[i-1][j-1]
        dp[i][j] = max(dp[i][j],dp[i-1][j],dp[i][j-1])
        if lst1[i] == lst2[j] : dp[i][j] = max(dp[i][j],dp[i][j-1]+1)
 
print(dp[M][L])