LCA问题的跳表算法 >> NoAlGo博客


LCA问题的跳表算法 » NoAlGo博客
LCA问题的一般形式:给定一棵有根树,给出若干个查询,每个查询要求指定节点u和v的最近公共祖先。
LCA问题有两类解决思路:
  • 在线算法,每次读入一个查询,处理这个查询,给出答案。
  • 离线算法,一次性读入所有查询,统一进行处理,给出所有答案
lca
跳表算法是在线算法,其需要O(nlogn)的复杂度进行初始化,然后每进行一个查询需要O(logn)复杂度进行解答。
使用跳表解决LCA问题时,对每个节点i维护了一个数组anc[i][j],表示i的第2^j个祖先的节点。有了这个数组后,节点x沿着树往上跳2^k层之后,就变成了节点anc[x][k]。当待查询的两个节点u和v在同一层时,我们把uv每次向上跳2的幂次的距离,最后同时到达了u和v的LCA节点。
可以知道,由于每次跳的距离都是2的幂,如果把uv和其LCA节点间的深度差用二进制表示,则其中的每个出现1的位置就是uv每次所要跳的距离。比如,要查找5和6的LCA,则把5和6都往上跳2^1层,同时到达0,0即为所求LCA。
但如果0上面还有很多节点,则5和6同时往上跳比2更大的距离(如4)时,也会到达同样的节点,但并不是LCA(不是最近)。
为了避免这个问题,我们在跳的时候不要让他们到达相同的节点,这样就不会跳过头了。最终u和v会跳到其LCA的下一层,u和v的父亲就是他们的LCA。即,查找5和6的LCA时,5最终会跳到1,6会跳到3,它们的父亲0为要求的LCA。
当待查询的两个节点u和v不在同一层时,假如u的层数比较深。这是我们先把u跳到和v同一层的祖先中,然后再使用上述相同层的方法。把u每次往上跳2的幂次的层数,如果跳了之后在v的上方了,则不跳,否则就跳。可以知道,将u和v之间的深度差用二进制表示,则其中每个出现1的位置就是要跳的距离。比如,要查找7和6的LCA,首先把7往上跳到和6同一层的5中,然后再使用上述方法求5和6的LCA即可。

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