light oj 1422 - Halloween Costumes （區間dp）

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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 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

 

題意：

給你n天需要穿的衣服的樣式，每次可以套著穿衣服，脫掉的衣服就不能再用了（可以再穿，但算cost），問至少要帶多少條衣服才能參加所有宴會。

思路：

很明顯為區間dp，dp[i][j]表示i~j天所需的最小數量。

考慮第j天穿不穿，如果穿的話那麼 dp[i][j]=dp[i][j-1]+1;

如果不穿的話，那麼需要有一個 k (i<=k<j)，第j天和第k天穿的衣服相同，將k+1~j-1衣服套著穿後全部脫掉，那麼

dp[i][j]=dp[i][k]+dp[k+1][j-1]；

這個問題就愉快的被我們解決了，區間dp，走起！

代碼：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#define maxn 105#define MAXN 100005#define mod 100000000#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,ans,cnt,tot;int a[maxn],dp[maxn][maxn];int main(){    int i,j,t,test=0;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++) dp[i][i]=1;        for(j=2;j<=n;j++)        {            for(i=1;i<j;i++)            {                dp[i][j]=dp[i][j-1]+1;                for(int k=i;k<j;k++)                {                    if(a[k]==a[j]) dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j-1]);                }            }        }        ans=dp[1][n];        printf("Case %d: %d\n",++test,ans);    }    return 0;}