Lightoj 1422(58/600)

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 ith integer ci (1 ≤ ci ≤ 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

这种题看着数据量肯定不是贪心
贪心一会也会发现做不出来
然后试试二分、
发现枚举答案以后和原来没什么区别
所以开始dp
这种东西的复杂度对于状态压缩来说太多了
多余线性或者各种优化太少了
很明显是区间dp
然后模拟一下状态
枚举一下区间
只有在不一样的时候+1
很舒服

#include<bits/stdc++.h>
using namespace std;
int tu[101],dp[101][101];
int u=0;
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        memset(dp,0,sizeof dp);
        for(int a=1;a<=n;a++)scanf("%d",&tu[a]);
        for(int a=1;a<=n;a++)dp[a][a]=1;
        for(int a=1;a<=n;a++)
        {
            for(int b=a-1;b>=1;b--)
            {
                dp[b][a]=0x3f3f3f3f;
                for(int c=b;c<a;c++)
                {
                    if(tu[a]==tu[c])
                    {
                        dp[b][a]=min(dp[b][a],dp[b][c]+dp[c+1][a-1]);
                    }
                    else dp[b][a]=min(dp[b][a],dp[b][c]+dp[c+1][a-1]+1);
                }
            }
        }
        printf("Case %d: %d\n",++u,dp[1][n]);
    }
}