LightOJ 1422 Halloween Costumes

Gappu has a very busy weekend ahead of him. Because, next weekend is Halloween, and he is planning to attend as many parties as he can. Since it's Halloween, these parties are all costume parties, Gappu always selects his costumes in such a way that it blends with his friends, that is, when he is attending the party, arranged by his comic-book-fan friends, he will go with the costume of Superman, but when the party is arranged contest-buddies, he would go with the costume of 'Chinese Postman'.

Since he is going to attend a number of parties on the Halloween night, and wear costumes accordingly, he will be changing his costumes a number of times. So, to make things a little easier, he may put on costumes one over another (that is he may wear the uniform for the postman, over the superman costume). Before each party he can take off some of the costumes, or wear a new one. That is, if he is wearing the Postman uniform over the Superman costume, and wants to go to a party in Superman costume, he can take off the Postman uniform, or he can wear a new Superman uniform. But, keep in mind that, Gappu doesn't like to wear dresses without cleaning them first, so, after taking off the Postman uniform, he cannot use that again in the Halloween night, if he needs the Postman costume again, he will have to use a new one. He can take off any number of costumes, and if he takes off k of the costumes, that will be the last k ones (e.g. if he wears costume Abefore costume B, to take off A, first he has to remove B).

Given the parties and the costumes, find the minimum number of costumes Gappu will need in the Halloween night.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer N (1 ≤ N ≤ 100) denoting the number of parties. Next line containsN integers, where the i^th integer c_i (1 ≤ c_i ≤ 100) denotes the costume he will be wearing in party i. He will attend party 1 first, then party 2, and so on.

Output

For each case, print the case number and the minimum number of required costumes.

Sample Input

2

4

1 2 1 2

7

1 2 1 1 3 2 1

Sample Output

            Case 1: 3

            Case 2: 4

给出n天的数字，每天需要穿相应数字衣服，可以套很多件在身上，脱掉就没了，问最少需要多少件衣服

初始化dp[i][i]应该都是1，但不是最优情况。 开始从小到大优化答案；

如果a[i]=a[j] 那么dp[i][j]=dp[i][j-1] 就是第i天和第j天的衣服相同，那么i衣服必定穿在身上，那么第j天就不需要再加衣服了；

再从中间找中间点来优化答案，从i到j天的衣服数量等于从i到k天+从k+1到j天的数量，根据这个来取最优；

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
	int t;
	cin>>t;
	int cs=0;
	while(t--)
	{
        int a[111],dp[111][111];
        int i,j;
        int n;
        cin>>n;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        cin>>a[i];
        for(i=1;i<=n;i++)
        dp[i][i]=1;
        for(j=1;j<=n;j++)
        for(i=1;j+i<=n;i++)
        {
        	int k=i+j;
        	dp[i][k]=99999999;
        	if(a[i]==a[k])dp[i][k]=dp[i][k-1];
        	for(int s=i;s<k;s++)
        	if(a[i]==a[s])dp[i][k]=min(dp[i][k],dp[i][s]+dp[s+1][k]);
		}
	 printf("Case %d: %d\n",++cs,dp[1][n]);
	}
	return 0;
}