— LCA, 图论, tarjan, 倍增 — 2 min read
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 17734 Accepted: 9405
A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:
In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.
For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest common ancestor of y and z is y.
Write a program that finds the nearest common ancestor of two distinct nodes in a tree.
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,..., N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.
Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.
2 16 1 14 8 5 10 16 5 9 4 6 8 4 4 10 1 13 6 15 10 11 6 7 10 2 16 3 8 1 16 12 16 7 5 2 3 3 4 3 1 1 5 3 5
4 3
Taejon 2002
tarjan+并查集求LCA模板。将u的子儿子v递归合并到u，若y在x的子树上，则根据并查集的性质，公共祖先即为x（father[y]）。若y和x不再同一子树上。则根据dfs回溯的性质，则搜完y之后一定是回溯到x,y的最近公共祖先才能搜到x的，而此时由并查集的性质，回溯的节点刚好就是father[y]。
LCA还可以用倍增来求：（思想很普通，只是以2进制的步长向上走：dp[i][j]表示节点i向上走2^j步到达的节点。）
LCA还可以转化为RMQ问题（个人更喜欢此方法，在线，且效率高）