状态压缩---区间dp第一题

标签： ACM

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题目

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

一道区间dp题，看得我头有点晕
题意为给你几个需要穿的衣服编号你需要依次穿上，可以套着穿，但是拖了不能再穿同一件（要增加次数），求最少的衣服数量
从最后一天往第一天dp，可以选择直接套上则dp[i][j]=dp[i+1][j]+1,如果中间有相同一套的衣服，可以选择把衣服脱了，则dp[i][j]=dp[i+1][k]+dp[k+1][j],再对这两个值取最小值即为dp[i][j]最小值

AC代码

#include <iostream>
#include <string.h>
using namespace std;
int dp[105][105];
int dress[105];
int main()
{
    int t,n,i,j,k,x;
    while(cin>>t)
    for(i=1;i<=t;i++)
    {
        memset(dp,0,sizeof(dp));
        cin>>n;
        for(j=0;j<n;j++)
            cin>>dress[j];
        for(j=0;j<n;j++)
            dp[j][j]=1;
        for(j=n-1;j>=0;j--)
            for(k=j+1;k<n;k++)
        {
            dp[j][k]=dp[j+1][k]+1;
            for(x=j+1;x<=k;x++)
                if(dress[x]==dress[j])
                dp[j][k]=min(dp[j][k],dp[j+1][x]+dp[x+1][k]);
        }
        cout<<"Case "<<i<<": "<<dp[0][n-1]<<endl;
    }
    return 0;
}