Lightoj 1422 Halloween Costumes(区间DP)

1422 - 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

 

PROBLEM SETTER: MANZURUR RAHMAN KHAN

SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

题目： http://lightoj.com/volume_showproblem.php?problem=1422

题意：给你n天需要穿的衣服的样式，每次可以套着穿衣服，脱掉的衣服就不能再穿了，问至少要带多少条衣服才能参加所有宴会

分析：把相同的衣服的那些天连一条边，说明，这两天可以只用一件衣服，然后发现，相交的连线有冲突，只能选择一条，这个问题就变成问，这些连线最多能选中几条。。。

注意，如果连线时包含的可以同时选中，一开始我忽略这个问题了，把它当做巨水的题= =

假设f[ i ][ j ]为区间[ i ] [ j ]最多能选中的连线，那么有 f[ i ] [ j ]=max{f[ i+1 ][ j]  ，  f[ i ] [ j-1 ]  ，  f[ i+1 ] [ j-1 ]+1(满足 i 和 j 相同的情况，也就是一条连线)  ，f[ i ][ k ]+f[ k ][ j ]} i<=k<=j

然后为了方便，直接记忆化搜素即可

PS：好吧，不知道什么叫区间DP，看有大牛的总结，好奇加担心，所以做一下，貌似以前做过类似的，不过没有模块化训练，我也就这么叫它吧= =

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int mm=222;
int f[mm][mm],c[mm];
int i,n,t,cs=0;
int DP(int i,int j)
{
    if(i>j)return 0;
    if(f[i][j]>-1)return f[i][j];
    f[i][j]=max(DP(i+1,j),DP(i,j-1));
    if(c[i]==c[j])f[i][j]=max(f[i][j],DP(i+1,j-1)+1);
    for(int k=i;k<=j;++k)
        f[i][j]=max(f[i][j],f[i][k]+f[k][j]);
    return f[i][j];
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(f,-1,sizeof(f));
        for(i=1;i<=n;++i)
            scanf("%d",&c[i]),f[i][i]=0;
        printf("Case %d: %d\n",++cs,n-DP(1,n));
    }
    return 0;
}