Light OJ 1422 Halloween Costumes(区间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 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

Output for Sample Input

2

4

1 2 1 2

7

1 2 1 1 3 2 1

Case 1: 3

Case 2: 4

 

题意：Gappu有n个parties要参加，每个party需要穿对应的服装，每个party之前Gappu可以穿上跟即将到来的party对应的衣服，或者脱掉若干衣服直到最上面的衣服跟party相对应，并且脱掉的衣服就不能在穿。

思路：第一道区间DP题，其他人的题解看了不理解，自己琢磨写写我的想法。我采用的也是从上到下更新，dp[i][j] 表示第i到第j个party需要最少的衣服数量，面临第i个party，我们必须穿上第i个party对应的衣服，那么dp[i][j] = dp[i + 1][j] + 1；如果第i + 1 到第j个party也有跟第i个party一样类型的party时，设其为第k个party，那么我们面临k时，可以把第i + 1到第k - 1个衣服脱掉，使得最上面的衣服跟第k个party相对应。所以dp[i][j] = min(dp[i + 1][j] + 1, dp[i + 1][k - 1] + dp[k][j])。

#include<cstdio>#include<algorithm>#include<map>#include<string>#include<iostream>#include<vector>#include<cmath>#include<ctime>#include<set>using namespace std;int a[110], dp[110][110];int main(){    int t;    scanf("%d", &t);    int ca = 1;    while(t--)    {        int n;        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 i = n - 1; i >= 1; i--)        {            for(int j = i + 1; j <= n; j++)            {                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: %d\n", ca++, dp[1][n]);    }    return 0;}