How to find the total number of Increasing sub-sequences of certain length with Binary Index Tree(BIT)

后端 未结 1 1399
一向
一向 2021-02-02 03:54

How can I find the total number of Increasing sub-sequences of certain length with Binary Index Tree(BIT)?

Actually this is a problem from Spoj Online Judge

相关标签:
1条回答
  • 2021-02-02 04:51

    Let:

    dp[i, j] = number of increasing subsequences of length j that end at i
    

    An easy solution is in O(n^2 * k):

    for i = 1 to n do
      dp[i, 1] = 1
    
    for i = 1 to n do
      for j = 1 to i - 1 do
        if array[i] > array[j]
          for p = 2 to k do
            dp[i, p] += dp[j, p - 1]
    

    The answer is dp[1, k] + dp[2, k] + ... + dp[n, k].

    Now, this works, but it is inefficient for your given constraints, since n can go up to 10000. k is small enough, so we should try to find a way to get rid of an n.

    Let's try another approach. We also have S - the upper bound on the values in our array. Let's try to find an algorithm in relation to this.

    dp[i, j] = same as before
    num[i] = how many subsequences that end with i (element, not index this time) 
             have a certain length
    
    for i = 1 to n do
      dp[i, 1] = 1
    
    for p = 2 to k do // for each length this time
      num = {0}
    
      for i = 2 to n do
        // note: dp[1, p > 1] = 0 
    
        // how many that end with the previous element
        // have length p - 1
        num[ array[i - 1] ] += dp[i - 1, p - 1]   
    
        // append the current element to all those smaller than it
        // that end an increasing subsequence of length p - 1,
        // creating an increasing subsequence of length p
        for j = 1 to array[i] - 1 do        
          dp[i, p] += num[j]
    

    This has complexity O(n * k * S), but we can reduce it to O(n * k * log S) quite easily. All we need is a data structure that lets us efficiently sum and update elements in a range: segment trees, binary indexed trees etc.

    0 讨论(0)
提交回复
热议问题