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By pimenta, 23 months ago, In English,

I was searching for this well-known technique, but I couldn't find any good tutorials about it, so I had to figure it out myself. And now I want to write my first contribution. The binary lifting method for answering LCA queries is explained in many places, like this, this and this (the method with cost ). I'll explain it one more time, emphasizing how it can be adapted to answer queries about paths on weighted trees (static version, so topology/weight updates are not supported).

Suppose we have a weighted tree G = (V, E), where V = {1, 2, ..., n} and is the weight function. First, we need to define some things, which can all be calculated with a simple DFS:

  1. p(u) is the parent of vertex u in the DFS spanning tree. p(r) = r, if r is the root.

  2. pw(u) is the weight of the edge from u to its parent, p(u). Therefore, pw(u) = w({u, p(u)}). If r is the root, pw(r) must be the identity element of your query, like for min, or  - ∞ for max.

  3. L(u) is the level of vertex u and is defined as L(u) = L(p(u)) + 1. L(r) = 0, if r is the root.

This method is interesting because it combines dynamic programming and divide and conquer paradigms. Let f(u, i) be the 2i-th ancestor of vertex u. So what is the recurrence relation? How can we break path from u to its 2i-th ancestor into pieces, so we can compute it more easily? Stop and think. If we're storing ancestors in powers of two, how about breaking the path in its half? Which vertex is exactly in the middle of u and its 2i-th ancestor? Yes, its 2i - 1-th ancestor. More precisely, f(u, i - 1). So the acyclic recurrence is

f(u, i) = f(f(u, i - 1), i - 1)

and the base case is

f(u, 0) = p(u)

The greatest i that makes sense for f(u, i) is such that 2i = O(n). So . Then we have DP states and computing one state costs O(1). Therefore, preprocessing will take up to operations (DFS + DP).

The intuitive algorithm to find LCA(u, v) using f(u, i) follows:

  1. Let r be the root. Choose u to be the farther vertex from r of u and v. In other words, swap u and v if L(v) > L(u).

  2. Let . Set u to be its ancestor in the same level of v, climbing up the tree. This can be done with , for i = k, k - 1, k - 2, ..., 0, if, at each iteration, f(u, i) is still not above v, that is, if L(f(u, i)) ≥ L(v).

  3. At this point, if u = v then LCA(u, v) = v and we can stop the algorithm.

  4. Finally, climb up the tree with u and v. In other words, make and , for i = k, k - 1, k - 2, ..., 0, if, at each iteration, f(u, i) and f(v, i) are still not above their LCA, that is, if f(u, i) ≠ f(v, i). At the end of this stage, p(u) = p(v) is the LCA we seek.

No doubt, this algorithm runs in time. Now, let's observe the strategy it is based on. First, we climb the path from u to its ancestor in the same level as v, then we climb from v and this new u to their LCA. Note that all these jumps split the path between original u and v vertices in disjoint paths. Therefore, we're actually dividing the problem in smaller disjoint subproblems. This is the principle of divide and conquer and all those algorithms and data structures for RMQ and LCA. So is easy to adapt the algorithm described above to answer queries about paths in the underlying tree. All we need is to define a function g(u, i) to give the optimal (or total) result for path from u to its 2i-th ancestor, which can be formally defined as

and

g(u, 0) = pw(u)

The adaptation to the 4-stage algorithm described above is very simple. Start the answer with the identity element of your query (like for min, or  - ∞ for max). Whenever you make or (stages 2 and 4), improve the answer with the respective result: g(u, i) or g(v, i). If stage 4 is actually executed, then is necessary to improve the answer with pw(u) and pw(v) in the end. That's all!

Thanks to dalex for the insight (here).

For more dynamic versions of this problem, we have:

  1. Heavy-light decomposition (here and here), so you can update weights.

  2. Link cut trees (here), so you can update both weights and topology.

Both HLD and LC trees can answer queries in time, but they're harder to code and understand.

 
 
 
 
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23 months ago, # |
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23 months ago, # |
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thank you! just in time

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23 months ago, # |
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USACO also provides a nice explanation (there was a LCA problem in the last (December) contest): http://usaco.org/current/data/sol_maxflow_platinum_dec15.html

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23 months ago, # |
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23 months ago, # |
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I know another way to do the LCA problem , this ideea just came in my mind because i misunderstood what you just wrote lol :)

Its Cool because it does not use use any data structure. and O(n) memory The ideea behind it : its a mix between a brute-force and a Fenwick tree ideea)

Here is the explanation

We want to Improve the Brute-Force ideea by keeping for each node an ancestor , which is at distance lsb(level[node]) upward ; where lsb(X) is the least semnificant bit ie (X&(X^(X-1));

check some google if you are not familiar with LSB(X) and Fenwick tree its a prety cool trick , check it it out :)

Keep a stack of nodes in dfs order . you will remove a node from the stack when u have visited all its subtree ; (when you are at node i you have the path from root to your node i)

In this way you can pre-calculate the ancestor you want to jump upwards to

Now i get a query node1,node2

Here i get check 3 cases

1: Can i jump lsb(level[node1]) , without node1 becoming the parent of node2? if yes them jump

2: Can i jump lsb(level[node2]) , without node2 becoming the parent of node1? if yes them jump

3: If none of the above are true i just move the lowest of the nodes 1 level upwards ( by moving to its direct parent)

repeat the above 3 steps until node1!=nod2

This works in log(N) since the complexity its very similar to binary indexed tree tehnique (also known as BIT or Fenwick Tree)

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    23 months ago, # ^ |
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    edit: this is NOT log2(N) but neither log2(N)^2 it behaves in the middle somewhere on my brute force generator

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22 months ago, # |
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12 months ago, # |
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Thank you, I have implemented in slightly different way http://pastebin.com/XMPmkb6W