如何用二叉索引树（BIT）求一定长度的递增子序列的总数

让：

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

一个简单的解决方案是：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]

答案是.dp[1, k] + dp[2, k] + ... + dp[n, k]

现在，这可行，但对于给定的约束来说效率低下，因为可以达到 . 足够小，所以我们应该尝试找到一种方法来摆脱.n10000kn

让我们尝试另一种方法。我们还有 - 数组中值的上限。让我们尝试找到与此相关的算法。S

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]

这很有复杂性，但我们可以很容易地将其简化。我们所需要的只是一个数据结构，让我们可以有效地对范围内的元素求和和更新：段树、二进制索引树等。O(n * k * S)O(n * k * log S)