Eva is trying to make her own color stripe（條紋） 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

?題目大意：從一串長度為L的序列中選出包含給定長度為M的子序列的最長序列(子序列元素的相對位置不能發生改變)，輸出其長度

?思路：1、將給定長度為M的子序列的位置編個號，從左往右編號遞增，要保證子序列元素的相對位置不發生改變，也就是說子序列元素的編號從左往右不下降，從而可以轉換成最長不下降子序列的問題（動態規劃-LIS）

2、轉換為最長不下降子序列問題：從長度為L的序列的序列中選出包含給定長度為M的子序列元素的元素，按照原來的順序存到A數組中，對A數組進行動態規劃

3、最長不下降子序列的問題（動態規劃-LIS）的狀態方程

（j=1,2,…,i-1&&A[j]<A[i]）

4注意編號用map可能會超時，可以自己用數組實現哈希編號

?AC代碼

#include <iostream> #include <vector> using namespace std; int main() {int n, m, k;scanf("%d", &n);scanf("%d", &m);int mp[10000] = {0};for (int i = 1; i <= m; i++) {int temp;scanf("%d", &temp); mp[temp] = i;} scanf("%d", &k);vector<int> A; for (int i = 0; i < k; i++) {int temp;scanf("%d", &temp); if (mp[temp]) {A.push_back(temp);}}vector<int> dp(A.size());int ans = -1;for (int i = 0; i < A.size(); i++) {dp[i] = 1;for (int j = 0; j < i; j++) {if (mp[A[i]] >= mp[A[j]] && dp[j] + 1 > dp[i]) {dp[i] = dp[j] + 1;}}ans =max(ans, dp[i]); }cout << ans;return 0; }

更多PAT甲級題目：請訪問PAT甲級刷題之路--題目索引+知識點分析（正在進行），感謝支持！

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