LightOJ 1422 Halloween Costumes

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探讨了在参加一系列万圣节派对时，如何通过合理的服装搭配和更换策略来减少所需服装数量的问题。采用区间动态规划算法求解，考虑了服装的叠加和去除规则。

Halloween Costumes

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1422

uDebug

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 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

1 2 1 2

1 2 1 1 3 2 1

Sample Output

Case 1: 3

Case 2: 4

题意：有n个宴会，每个宴会要穿指定衣服，衣服可以套着穿，也可以脱下，脱下的就不能再穿，问最少需要几件礼服。

思路：区间dp，dp[i][j]表示第i个宴会到第j个宴会所需要的最少礼服，有动态转移方程：dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j])(第i个宴会的礼服和第k个宴会的礼服相同)。下面给代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<bitset>
#include <utility>
using namespace std;
#define maxn 105
typedef long long LL;
const int mod = 1e8;
int a[maxn], dp[maxn][maxn];
int main(){
	int t;
	scanf("%d", &t);
	for (int tcase = 1; tcase <= t; tcase++){
		int n;
		scanf("%d", &n);
		for (int i = 0; i < n; i++){
			scanf("%d", &a[i]);
		}
		for (int i = 0; i < n; i++){
			dp[i][i] = 1;
		}
		for (int i = 1; i < n; i++){
			for (int j = 0; j + i < n; j++){
				dp[j][j + i] = dp[j + 1][j + i] + 1;
				for (int k = j + 1; k <= j + i; k++){
					if (a[j] == a[k]){
						dp[j][j + i] = min(dp[j][j + i], dp[j + 1][k - 1] + dp[k][j + i]);
					}
				}
			}
		}
		printf("Case %d: %d\n",tcase, dp[0][n - 1]);
	}
}