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};//記錄某一顏色是否是最喜歡的顏色 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)//只保留在最喜歡的顏色里出現(xiàn)過的顏色 		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--){//計算每個字符之后的字符串最大長度 		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++)//找到最大長度 		if(c[i]>max)		max=c[i];	cout<<max;} 感想：想了很久只想到這種方法，從后往前，每個符合要求的字符后的字符串長度是后面的字符符合要求的字符串的長度加一，這樣一直循環(huán)就可以得到最大值