LightOJ

Description

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 A before 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 contains N integers, where the ith integer ci(1 ≤ ci ≤ 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

题意：有N个宴会，对于每一个宴会，女猪脚都要穿一种礼服，礼服可以套着穿，但是脱了的不能再用，参加宴会必须按顺序来，从第一个到第N个，问参加这些宴会最少需要几件礼服

        比如说第一个样例：第一天穿衣服“1”，然后不脱，在第二天穿上第二件衣服“2”，第三天脱掉衣服“2”，第四天穿上新的衣服“2”，总共要3件衣服。

思路：区间dp，从后往前推

        首先每增加一天要增加一件衣服，dp[i][j]=dp[i+1][j]+1.

        然后k在i+1~j的区间，若a[i]==a[k]说明衣服可以不用脱下来，可以穿着在k天的时候再穿，所以只需要选一件即可。所有转移方程为：dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j])

        再看边界问题：有两种初始化的方法：

        1、for(int i=1;i<=n;i++){

               dp[i][i]=1;
           }即每一天需要一件衣服

        2、区间的DP值 可以设为 i,j的区间长度即可，

       for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
               dp[i][j] = j - i + 1;

AC代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 106
#define inf 1e12
int n;
int a[N],dp[N][N];
int main()
{
   int t;
   int ac=0;
   scanf("%d",&t);
   while(t--){
      scanf("%d",&n);
      for(int i=1;i<=n;i++){
         scanf("%d",&a[i]);
      }

      for(int i=1;i<=n;i++){
         dp[i][i]=1;
      }
      for(int len=1;len<n;len++){
         for(int i=1;i+len<=n;i++){
            int j=i+len;
            dp[i][j]=dp[i+1][j]+1;
            for(int k=i+1;k<=j;k++){
               if(a[i]==a[k]){
                  dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);
               }
            }
         }
      }
      printf("Case %d: ",++ac);
      printf("%d
",dp[1][n]);
   }
    return 0;
}