1045. Favorite Color Stripe (30)

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 (<=10000) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line are 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:

65 2 3 1 5 612 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7

#include<iostream>#include<algorithm>using namespace std; int color[201]={0};//记录某一颜色是否是最喜欢的颜色 int main(){	int N,M,i,j,temp,count=0;	cin>>N;	cin>>N;	int *a=new int[N];	for(i=0;i<N;i++){		scanf("%d",&a[i]);		color[a[i]]=++count;	}	count=0;	cin>>M;	int *b=new int[M];	for(i=0;i<M;i++){		scanf("%d",&temp);		if(color[temp]>0)//只保留在最喜欢的颜色里出现过的颜色 		b[count++]=temp;	}	M=count;	int *c=new int[M];	for(i=0;i<M;i++)	c[i]=1;	int max;	for(i=M-1;i>=0;i--){//计算每个字符之后的字符串最大长度 		max=0;	for(j=M-1;j>i;j--)		if(color[b[j]]>=color[b[i]])		if(max<c[j])		max=c[j];	c[i]=max+1;	}	for(i=0;i<M;i++)//找到最大长度 		if(c[i]>max)		max=c[i];	cout<<max;} 感想：想了很久只想到这种方法，从后往前，每个符合要求的字符后的字符串长度是后面的字符符合要求的字符串的长度加一，这样一直循环就可以得到最大值