[light1422][区间dp]Halloween Costumes

1422 - Halloween Costumes
PDF (English) Statistics Forum
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 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
Output for Sample Input
2
4
1 2 1 2
7
1 2 1 1 3 2 1
Case 1: 3
Case 2: 4

PROBLEM SETTER: MANZURUR RAHMAN KHAN
SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

区间dp：

区间dp是个好东西，如果看到一个n可以在n^3内解决问题，那么他基本上不是高消就是区间dp(有的数据范围是n^2,那个要用到一个叫做四边形不等式优化的东西)。区间dp和普通dp没有什么区别，如果硬要说区别的话，也许就是他的转移更加的有特点了。他的区间和区间之间往往有千丝万缕的联系，使得你设计的dp状态看似存在后效性，但是如果我们认真观察，就能发现这些状态之间的决策并不互相影响，决定能否转移的仅仅是给定的属性（输入的价值，给定的位置等等）。所以其实就是一个较为抽象的dp（其实我感觉抽象应该只是我做的不熟练）。

sol:

题意是给你n天要穿的衣服，衣服a可以套在衣服b外面，然后衣服b脱掉又会变成衣服a，b脱掉后可以再穿上（次数+1），一次可以在身上穿多个衣服a，问至少要带多少件衣服。
设dp[i][j]表示i到j天最少要穿几件衣服。那么区间l,r首先可以从l+1,r转移过来，我们直接让他在l天多穿一件衣服即可。还有就是如果在第k天(i< k<=j)穿了和第i天一样的衣服，我们就可以直接用第k天的衣服。但是要求是第k天的衣服必须不被脱下，所以i+1到k-1天的衣服就要独立出来决策。

许多区间dp其实都是差不多的，大多是通过一些方法来合并2个区间。

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
using namespace std;
int n,m;
inline int read()
{
    char c;
    int res,flag=0;
    while((c=getchar())>'9'||c<'0') if(c=='-')flag=1;
    res=c-'0';
    while((c=getchar())>='0'&&c<='9') res=(res<<3)+(res<<1)+c-'0';
    return flag?-res:res;
}
const int N=110;
int f[N][N],a[N];
inline void get(int l,int r)
{
    if(l>r) return;
    if(l==r)
    {
        f[l][r]=1;
        return;
    }
    if(f[l][r]) return;
    get(l+1,r);
    f[l][r]=f[l+1][r]+1;
    for(int k=l+1;k<=r;++k)
    if(a[k]==a[l])
    {
        get(l+1,k-1);
        get(k,r);
        f[l][r]=min(f[l][r],f[l+1][k-1]+f[k][r]);
    }
}
int cnt;
inline void solve()
{
    ++cnt;
    memset(f,0,sizeof(f));
    n=read();
    for(int i=1;i<=n;++i) a[i]=read();
    get(1,n);
    printf("Case %d: %d\n",cnt,f[1][n]);
}
int main()
{
//  freopen("1422.in","r",stdin);
//  freopen(".out","w",stdout);
    int T=read();
    while(T--) solve();
}