lightoj-1422 - Halloween Costumes 解题报告 区间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

题目大意，不同的聚会穿不同的衣服，数据结构是堆栈，脱下的衣服不能再穿上，求最少需要的衣服。

状态dp[i][j]表示从第i场聚会到第j场聚会需要穿的最少衣服，可以想到计算dp[i][j]是需要用到i之后的状态，因此i是倒着枚举的。

由于每场聚会至少穿一件衣服，所以初始化时dp[i][j]=1。

对于dp[i][j]，如果在其之后的某场k有c[i]==c[k]，那么则可以尝试不脱掉第i场的衣服，即dp[i][j]=min(dp[i][j], dp[i+1][k-1]+dp[k][j])，即所谓的区间dp。

而对于c[i]!=c[k]，则dp[i][j]=dp[i+1][j]+1。

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int n;int c[107];int dp[107][107];int t=1;void solve(){    for(int i=0;i<n;i++){              //初始化        for(int j=i;j<n;j++){            dp[i][j]=1;        }    }    for(int i=n-2;i>=0;i--){        for(int j=i+1;j<n;j++){            dp[i][j]=dp[i+1][j]+1;       //先假设c[i]!=c[j]            for(int k=i+1;k<=j;k++){                if(c[i]==c[k])                               dp[i][j]=min(dp[i][j], dp[i+1][k-1]+dp[k][j]);            }        }    }    printf("Case %d: %d\n", t++, dp[0][n-1]);}int main(){    int T;    scanf("%d", &T);    while(T--){        scanf("%d", &n);        memset(dp, 0, sizeof(dp));        for(int i=0;i<n;i++){            scanf("%d", &c[i]);        }        solve();    }}