1045 Favorite Color Stripe (30 point(s))

1045 Favorite Color Stripe (30 point(s))

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 (≤10​4​​) 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

本质：动态规划。基于优先级的最长上升子串（LIS）。

相关链接：https://blog.youkuaiyun.com/coderwait/article/details/89679684

根据输入的favorite order进行优先级赋值，即先输入的优先级更高，后输入的优先级更低，优先级更低的只能出现在优先高的之后，不在favorite order中的不纳入考虑。

然后遍历每个元素，如果它在favorite order中，则再遍历之前的元素，如果priority[a[j]]<=priority[a[i]]则更新dp[]。dp[i]表示以第i个元素结尾的最长的满足要求的子串。最终的最长子串即为dp[]的最大值。

#include<iostream>
#include<cstring>
using namespace std;
const int MAX = 1e4+7;
int priority[207]={0};
int a[MAX];
int dp[MAX]={0};//以位置i结束的最长串 
int main(void){
	int N,M,L,x;cin>>N>>M;//total colors, favorite order
	for(int i=1;i<=M;i++){
		cin>>x;
		priority[x]=i;//优先级，正数且优先级越小越高 
	}
	cin>>L; 
	for(int i=0;i<L;i++) cin>>a[i];
	int maxLen=0; 
	for(int i=0;i<L;i++){
		if(priority[a[i]]>0){
			for(int j=0;j<i;j++){
				if(priority[a[j]]>0&&priority[a[j]]<=priority[a[i]]){
					dp[i]=max(dp[i],dp[j]+1);
				}
			}
			dp[i]=max(1,dp[i]);
			maxLen = max(maxLen,dp[i]);
		}
	} 
	cout<<maxLen<<endl; 
	return 0;
}