LightOJ - 1422 Halloween Costumes(区间DP模板)

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 offA, 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 Nintegers, 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

题目大意:给你n个数，每个数意味着一件衣服，你可以套多件衣服，但是一旦脱下来某件衣服，之后要是遇到这件衣服，必须重新穿一个新的，一共至少穿几个衣服

解题思路:这道题是个区间DP，区间DP一般有三层循环，第一层循环是区间的长度，第二层循环是区间头尾位置，第三层循环是在区间中找个中间值比较求极值

对于这道题来说，dp[j][k]表示j到k区间一共至少穿几件衣服，如果dp[j]==dp[k]意味着他就不需要再穿一件衣服了，所以是等于dp[j][k-1]

然后比较一下到底dp[j][k]区间极值

#include<iostream>    
#include<cstdio>  
#include<stdio.h>  
#include<cstring>    
#include<cstdio>    
#include<climits>    
#include<cmath>   
#include<vector>  
#include <bitset>  
#include<algorithm>    
#include <queue>  
#include<map>  
using namespace std;

int main()
{
	int a[105], dp[105][105];
	int T, tot, i, n, k, j, u;
	cin >> T;
	tot = 0;
	while (T--)
	{
		tot++;
		cin >> n;
		memset(a, 0, sizeof(a));
		memset(dp, 0, sizeof(dp));
		for (i = 1; i <= n; i++)
		{
			dp[i][i] = 1;
		}
		for (i = 1; i <= n; i++)
		{
			cin >> a[i];
		}
		for (i = 2; i <= n; i++)
		{
			for (j = 1,k = i; k <=n; j++, k++)
			{
				dp[j][k] = 100000;
				if (a[j] == a[k])
				{
					dp[j][k] = dp[j][k - 1];
				}
				for (u = j; u < k; u++)
				{
					dp[j][k] = min(dp[j][k], dp[j][u] + dp[u + 1][k]);
				}
			}
		}
		cout << "Case " << tot << ": " << dp[1][n] << endl;
	}
}