1045 Favorite Color Stripe（30分）-PAT甲级

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

 题目大意：从一串长度为L的序列中选出包含给定长度为M的子序列的最长序列(子序列元素的相对位置不能发生改变)，输出其长度

 思路：1、将给定长度为M的子序列的位置编个号，从左往右编号递增，要保证子序列元素的相对位置不发生改变，也就是说子序列元素的编号从左往右不下降，从而可以转换成最长不下降子序列的问题（动态规划-LIS）

2、转换为最长不下降子序列问题：从长度为L的序列的序列中选出包含给定长度为M的子序列元素的元素，按照原来的顺序存到A数组中，对A数组进行动态规划

3、最长不下降子序列的问题（动态规划-LIS）的状态方程

（j=1,2,…,i-1&&A[j]

4注意编号用map可能会超时，可以自己用数组实现哈希编号

 AC代码

#include 
#include 
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 A; 
	for (int i = 0; i < k; i++) {
		int temp;
		scanf("%d", &temp); 
		if (mp[temp]) {
			A.push_back(temp);
		}
	}
		vector 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;
}

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