TopCoder TaroCards

https://community.topcoder.com/stat?c=problem_statement&pm=13015&rd=15846

Cat Taro has N cards. Exactly two integers are written on each card. You are given two int[]s first and second, each with N elements. For each i, the element first[i] represents the first integer on the i-th card, and the element second[i] represents the second integer on the i-th card. It is known that for each x from 1 to N, inclusive, there is exactly one card with the first integer equal to x. In other words, all elements of first represent a permutation of integers from 1 to N, inclusive. On the other hand, second may contain duplicates, but all elements of second are only between 1 and 10, inclusive. You are also given an int K. Taro wants to choose some subset of the cards (possibly none or all of them) in such a way that the total number of different integers written on the cards is less than or equal to K. Return the total number of ways to do that.   Definition      Class: TaroCards Method: getNumber Parameters: int[], int[], int Returns: long Method signature: long getNumber(int[] first, int[] second, int K) (be sure your method is public)        Constraints - first will contain between 1 and 50 elements, inclusive. - first and second will contain the same number of elements. - first will represent a permutation of integers between 1 and N, inclusive, where N is the number of elements in first. - Each element of second will be between 1 and 10, inclusive. - K will be between 1 and 2N, inclusive, where N is the number of elements in first.   Examples 0)      {1, 2} {2, 3} 2 Returns: 3 In this case, there are four subsets of cards: None of the cards. The number of different integers is 0. Only the first card. The number of different integers is 2. Only the second card. The number of different integers is 2. Both the first and the second card. The number of different integers is 3. However, the last subset has too many different integers. Thus, the answer is 3. 1)      {3, 1, 2} {1, 1, 1} 3 Returns: 8 2)      {4, 2, 1, 3} {1, 2, 3, 4} 5 Returns: 16 3)      {1, 2, 3, 4, 5, 6, 7} {2, 1, 10, 9, 3, 2, 2} 3 Returns: 17 4)      {1} {2} 1 Returns: 1 5)      {6, 20, 1, 11, 19, 14, 2, 8, 15, 21, 9, 10, 4, 16, 12, 17, 13, 22, 7, 18, 3, 5} {4, 5, 10, 7, 6, 2, 1, 10, 10, 7, 9, 4, 5, 9, 5, 10, 10, 3, 6, 6, 4, 4} 14 Returns: 2239000 This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2010, TopCoder, Inc. All rights reserved.

https://codesea.wordpress.com/2014/03/31/a-easy-dp-problem-with-bitmask/

Given N integers. Let Total subsets T=2^N. How many subsets are available from T where the distinct integers are less than or equal to K.

LL dp[59][1055];

LL a[59],n,k;

 

LL f(LL i,LL mask)

{

    if(dp[i][mask]!=-1) return dp[i][mask];

    if(i==n)

    {

        LL p=__builtin_popcount(mask);

        if(p<=k) return 1;

        else return 0;

    }

    LL res=f(i+1,mask); //not using i-th integer

    LL nmask=mask;

    nmask=nmask|(1<<a[i]);

    res=res+f(i+1,nmask); //using i-th integer

    return dp[i][mask]=res;

}

 

 

int main()

{

    LL i,j,m,d;

    while(cin>>n>>k)

    {

        memset(dp,-1,sizeof(dp));

        for(i=0;i<n;i++)

            cin>>a[i];

        LL ans=f(0,0);

        cout<<ans<<endl;

    }

    return 0;

}

https://blog.csdn.net/stafuc/article/details/26636451
枚举每张卡片选或不选，复杂度250 ，考虑到限制1-10的数字可能重复出现，11-50的数字只会出现一次。因此在250 枚举中有很多重复的状态，把计算结果保留下来，使每种状态只计算一次，降低复杂度。每次枚举时保存一个值记录1-10的数字都有哪些出现过，另保存一个值记录11-50中的数字已经出现多少个了。


https://github.com/livingstonese/topcoder/blob/master/src/main/java/srm613/division2/TaroCards.java

dp[i][sta][k]表示枚举到第i张卡，<=10的数字状态为sta，共有k个不同的数字的方案数，转移分第i张卡取或者不取

  private static final int MAX_CARD_SET = 1024;

  private static final int MAX_SINGLE_CARD_INTEGERS = 41;

  private static final int MAX_CARDS = 51;

  long[][][] dp = new long[MAX_CARD_SET][MAX_SINGLE_CARD_INTEGERS][MAX_CARDS];

  boolean[][][] visited = new boolean[MAX_CARD_SET][MAX_SINGLE_CARD_INTEGERS][MAX_CARDS];

  public long getNumber(int[] first, int[] second, int K) {

    return getNumberUtil(first, second, K, 0, 0, 0);

  }

  private long getNumberUtil(int[] first, int[] second, int K, int S, int r, int i) {

    if (visited[S][r][i])

      return dp[S][r][i];

    if (i == first.length) {

      dp[S][r][i] = (Integer.bitCount(S) + r) <= K ? 1 : 0;

      visited[S][r][i] = true;

      return dp[S][r][i];

    }

    visited[S][r][i] = true;

    // Don't pick card i

    dp[S][r][i] = getNumberUtil(first, second, K, S, r, i+1);

    // Pick card i

    int S1 = S, r1=r;

    

    if (first[i] < 11)

      S1 = S1 | (1 << (first[i]-1));

    else

      r1++;

    S1 = S1 | (1 << (second[i]-1));

    dp[S][r][i] += getNumberUtil(first, second, K, S1, r1, i+1);

    return dp[S][r][i];

  }

https://codesea.wordpress.com/2014/03/31/a-easy-dp-problem-with-bitmask/

LL dp[59][1055];

LL a[59],n,k;

 

LL f(LL i,LL mask)

{

    if(dp[i][mask]!=-1) return dp[i][mask];

    if(i==n)

    {

        LL p=__builtin_popcount(mask);

        if(p<=k) return 1;

        else return 0;

    }

    LL res=f(i+1,mask); //not using i-th integer

    LL nmask=mask;

    nmask=nmask|(1<<a[i]);

    res=res+f(i+1,nmask); //using i-th integer

    return dp[i][mask]=res;

}

 

 

int main()

{

    LL i,j,m,d;

    while(cin>>n>>k)

    {

        memset(dp,-1,sizeof(dp));

        for(i=0;i<n;i++)

            cin>>a[i];

        LL ans=f(0,0);

        cout<<ans<<endl;

    }

    return 0;

}