Lightoj 1422 区间dp

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

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

题目大意：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

我的贪心错误代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <cmath>using namespace std;int a[105];int b[105];int dp[105];int main(){    int T;    int tt=1;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        memset(b,0,sizeof(b));        dp[0]=1;        b[a[0]]=1;        for(int i=1;i<n;i++)        {            if(b[a[i]]==0)            {                dp[i]=dp[i-1]+1;                b[a[i]]=1;            }            else            {                dp[i]=dp[i-1];                for(int j=i-1;a[j]!=a[i];j--)                {                    b[a[j]]=0;                }            }        }        printf("Case %d: %d\n",tt++,dp[n-1]);    }    return 0;}

数据：

6

1 2 3 1 3 2   就过不了orz

其实这是一道区间dp的题目：dp[i][j]表示在区间i~j内最少要穿的衣服的件数。

思路：我们从后往前推导，dp[i][j]代表从区间i到区间j最少的穿衣数量，那么在dp[i][j]这个状态的穿衣数，

就要等于dp[i+1][j]+1；也就是说，首先在不考虑它后面是否有一天要穿相同的衣服的情况下，它肯定会比区间i+1到j的衣服多出一件；
然后，再考虑在这个区间范围，是否有一天要穿相同的衣服，i<k<=j，如果有第k天衣服和第i天的衣服是一样的，那么就要比较如果第i天不穿1件衣服与第i天穿上1件衣服；
首先，第i天穿上一件衣服的结果已经得出，那么我们只需比较不穿衣服，那么就是dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);

#include <string.h>#include <stdio.h>#include <iostream>using namespace std;const int N=105;int a[N],dp[N][N];int main(){    int T,tt=1;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            for(int j=i;j<=n;j++)                  dp[i][j]=j-i+1;        for(int i=n-1;i>0;i--)            for(int j=i+1;j<=n;j++)            {                dp[i][j]=dp[i+1][j]+1;                for(int k=i+1;k<=j;k++)                {                    if(a[i]==a[k])                        dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k][j]);                }            }        printf("Case %d: %d\n",tt++,dp[1][n]);    }    return 0;}