LightOJ1422-Halloween Costumes

                                                                                                                   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：告诉有n场晚会中需要穿的衣服，衣服是可以套在其他衣服外面的，告诉了序列顺序之后求出最少需要穿多少次衣服。

解题思路：使用dp[i][j]来表示区间 i~j的答案，那么对于第 i 件衣服，我们有

①：如果在之后的区间内都不再重复利用这件衣服，那么明显  dp[i][j] = dp[i+1][j] + 1

②：如果在之后的区间 i+1 ~ j 中存在一件衣服 k 是跟 i 一样的，那么我们便可以考虑是不是可以将i那件衣服在k这个地方重复利用，那么转移方程为  dp[i][j] = min(dp[i][j] , dp[i][k-1]+dp[k+1][j])

因为是从后面的区间往前面转，所以循环的时候应该是从后往前的

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;int main(){    int t,a[105],dp[105][105],n,cas=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++)            for(int j=i; j<=n; j++)                dp[i][j]=j-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[k]==a[i])                        dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]);                }            }        }        printf("Case %d: %d\n",++cas,dp[1][n]);    }    return 0;}