最低的Common Ancestor动态编程方法

Question

我正在学习LCA并遇到了Tutorial。我无法理解它的O（log n）方法。

首先，让我们计算一个表P[1,N][1,logN]，其中P[i][j]是2^j - i的祖先。为了计算这个值，我们可以使用以下递归：

预处理功能

void process3(int N, int T[MAXN], int P[MAXN][LOGMAXN])
  {
      int i, j;

  //we initialize every element in P with -1
      for (i = 0; i < N; i++)
          for (j = 0; 1 << j < N; j++)
              P[i][j] = -1;

  //the first ancestor of every node i is T[i]
      for (i = 0; i < N; i++)
          P[i][0] = T[i];

  //bottom up dynamic programing
      for (j = 1; 1 << j < N; j++)
         for (i = 0; i < N; i++)
             if (P[i][j - 1] != -1)
                 P[i][j] = P[P[i][j - 1]][j - 1];
  }

查询功能

int query(int N, int P[MAXN][LOGMAXN], int T[MAXN], 
  int L[MAXN], int p, int q)
  {
      int tmp, log, i;

  //if p is situated on a higher level than q then we swap them
      if (L[p] < L[q])
          tmp = p, p = q, q = tmp;

  //we compute the value of [log(L[p)]
      for (log = 1; 1 << log <= L[p]; log++);
      log--;

  //we find the ancestor of node p situated on the same level
  //with q using the values in P
      for (i = log; i >= 0; i--)
          if (L[p] - (1 << i) >= L[q])
              p = P[p][i];

      if (p == q)
          return p;

  //we compute LCA(p, q) using the values in P
      for (i = log; i >= 0; i--)
          if (P[p][i] != -1 && P[p][i] != P[q][i])
              p = P[p][i], q = P[q][i];

      return T[p];
  }

请帮助我理解概念以及如何获得递归公式P[P[i][j-1][j-1]。