Halloween Costumes （LightOJ 1422）

Halloween Costumes

Time Limit: 2000MSMemory Limit: 32768KB64bit IO Format: %lld & %llu

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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

2

4

1 2 1 2

7

1 2 1 1 3 2 1

Sample Output

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，也就是说如果顺序为 1 2 1，那么可以将2套在1外面，第三场晚会需要穿1的时候把2脱掉即可，这样就只需要穿两次衣服。

题目是再告诉了顺序之后需要求出在某种序列下最少需要穿多少次衣服。

样例 1 2 1 2： ①穿1   ②穿2  ③脱2  ④穿2      或者     ①穿1   ②脱1穿2  ③穿1  ④脱1    均输出  3

思路：

感觉思路比较巧妙，首先我们使用dp[a][b]来表示区间 a~b 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]);

或者认为k那件衣服是来自于i 转移方程为    dp[i][j] = min(dp[i][j] , dp[i+1][k-1]+dp[k][j]); //这个转移方程和上一个同意，可以任选一个。

还有一个要点就是，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"cstring"#include"cstdio"using namespace std;int n;int a[105];int dp[105][105];int main(void){    int t;    int cas = 0;    scanf("%d",&t);    while(t--)    {        cas ++;        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[i] == a[k])                    {                        dp[i][j] = min(dp[i][j],dp[i][k-1] + dp[k+1][j]);                        //dp[i][j] = min(dp[i][j],dp[i+1][k-1] + dp[k][j]);    用这个也可以                    }                }            }        }        printf("Case %d: %d\n",cas,dp[1][n]);    }    return 0;}