light oj 1422 Halloween Costumes（區間dp）

1422 - Halloween Costumes

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
2 4 1 2 1 2 7 1 2 1 1 3 2 1	Case 1: 3 Case 2: 4

參加n個party 參加特定的party需要穿特定的衣服  每次穿的衣服 可以套在之前的衣服上 或者把外面的衣服脫掉  但是脫掉的不能再穿了。。問最少需要多少件衣服

dp[i][j]表示第i天與第j天之間最少需要的衣服數量 對于第j件選擇是穿上或者是脫掉

如果第k件與第j件相同 那麼就取這兩種情況的最小值

如果第k件穿上 第j件不穿 就是把第k+1到第j-1的脫掉 是以dp[i][j]=dp[i][k]+dp[k+1][j-1]

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int a[110];
int dp[110][110];

int main()
{
//    fread;
    int cs=1;
    int tc;
    scanf("%d",&tc);
    while(tc--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        MEM(dp,0);
        for(int i=1;i<=n;i++)
            dp[i][i]=1;
        for(int j=2;j<=n;j++)
        {
            for(int i=1;i<j;i++)
            {
                dp[i][j]=dp[i][j-1]+1;
                for(int k=i;k<j;k++)
                {
                    if(a[k]==a[j])
                        dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);
                }
            }
        }
        printf("Case %d: %d\n",cs++,dp[1][n]);
    }
    return 0;
}