如何找到增加一定长度的子序列与二进制树索引的总数（BIT） [英] How to find the total number of Increasing sub-sequences of certain length with Binary Index Tree(BIT)

问题描述

其实这是一个 SPOJ 的问题。

假设我有一个数组 1,2,2,10 。

日益子序列长度为3的是 1,2,4 和 1,3,4 （基础）的索引。

The increasing sub-sequences of length 3 are 1,2,4 and 1,3,4(index based).

那么，答案是 2 。

推荐答案

让：

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

这是简单的解决方案是在为O（n ^ 2 * K）：

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]

答案是： DP [1，K] + DP [2，K] + ... + DP [N，K] 。

现在，这工作，但它是低效率的你给定约束，因为 N 可高达 10000 。 K 足够小，所以我们应该尽量找到一种方法来摆脱一个 N 。

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.

让我们尝试另一种方法。我们也有取值 - 上限的价值观在我们的数组。让我们试着找到与此相关的算法。

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]

这有复杂 O（N * K * S），但我们可以把它降低到 O（N * K *登录S）很容易。我们所需要的是一种数据结构，可以让我们更有效地在一个范围内总结和更新的元素：段树，<一个href="http://community.top$c$cr.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees">binary索引树等。

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.

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