PAT A1045 Favorite Color Stripe

1045 Favorite Color Stripe

分数 30

作者 CHEN, Yue

单位 浙江大学

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

 

* 最长不下降子串的模板；
 * 先对喜欢的数字赋一个喜欢度，后选择的数字的喜欢度必定要大于等于前面选择的数字的喜欢度；
 * 后面就是最长不下降子串的模板；
 *
 * 状态设计：dp[i] 表示所有以第i个数结尾的上升子序列的集合的最长长度；
 * 状态转移方程：dp[i] = max(dp[j]+1,1) (j=0,1,...,i-1);
 * dp[i] = dp[j] + 1 的含义表示i加到j的后面能否形成更长的LIS；

/*** 最长不下降子串的模板；* 先对喜欢的数字赋一个喜欢度，后选择的数字的喜欢度必定要大于等于前面选择的数字的喜欢度；* 后面就是最长不下降子串的模板；* * 状态设计：dp[i] 表示所有以第i个数结尾的上升子序列的集合的最长长度；* 状态转移方程：dp[i] = max(dp[j]+1,1) (j=0,1,...,i-1);* dp[i] = dp[j] + 1 的含义表示i加到j的后面能否形成更长的LIS；
*/#include <iostream>
#include <algorithm>using namespace std;const int N = 510, M = 1e5+10;
int dp[M], fav[M], ori[M];
int hs[N];int main()
{fill(hs, hs+N, -1); //初始化int n, m, l;cin >> n;cin >> m;for(int i=0; i<m; ++i)  {cin >> fav[i];hs[fav[i]] = i;}cin >> l;for(int i=0; i<l; ++i)  cin >> ori[i];for(int i=0; i<l; ++i){if(hs[ori[i]] != -1)    dp[i] = 1; //只有在喜欢的序列里面才可以选择该数字for(int j=0; j<i; ++j)if(hs[ori[i]] != -1 && hs[ori[i]] >= hs[ori[j]] && dp[j] + 1 > dp[i]){dp[i] = dp[j] + 1;}}cout << *max_element(dp, dp+M);return 0;
}