In a rooted tree, the lowest common ancestor (or LCA for short) of two vertices u and v is defined as the lowest vertex that is ancestor of both that two vertices.
Given a tree of N vertices, you need to answer the question of the form "r u v" which means if the root of the tree is at r then what is LCA of u and v.
Input
The first line contains a single integer N. Each line in the next N - 1 lines contains a pair of integer u andv representing a edge between this two vertices.
The next line contains a single integer Q which is the number of the queries. Each line in the next Q lines contains three integers r, u, v representing a query.
Output
For each query, write out the answer on a single line.
Constraints
20 points:
- 1 ≤ N, Q ≤ 100
40 points:
- 1 ≤ N, Q ≤ 105
- There is less than 10 unique value of r in all queries
40 points:
- 1 ≤ N, Q ≤ 2 × 105
Example
Input: 4 1 2 2 3 1 4 2 1 4 2 2 4 2 Output: 1 2Explanation
- "1 4 2": if 1 is the root, it is parent of both 2 and 4 so LCA of 2 and 4 is 1.
- "2 4 2": the root of the tree is at 2, according to the definition, LCA of any vertex with 2 is 2.
题意：给出一棵Ｎ个结点的树，有Ｑ次询问，每次询问给出三个数ｒ，ｘ，ｙ。
求当以ｒ作为树根时，ｘ和ｙ的lca
解决本题，有两个关键的地方：
１. 每次询问的答案只可能是: x, y, r, lca(x, y), lca(x, r), lca(y, r)，这里的lca都是以１为树根时的lca
2. 如果 x = lca(u, v), 那么dist(x, root) + dist(x, u) + dist(x, v)的值是最小的。
Accepted Code:
1 /*************************************************************************
2 > File Name: TALCA.cpp
3 > Author: Stomach_ache
4 > Mail: sudaweitong@gmail.com
5 > Created Time: 2014年09月24日 星期三 17时39分16秒
6 > Propose:
7 ************************************************************************/
8 #include <cmath>
9 #include <string>
10 #include <cstdio>
11 #include <vector>
12 #include <fstream>
13 #include <cstring>
14 #include <iostream>
15 #include <algorithm>
16 using namespace std;
17 /*Let's fight!!!*/
18
19 const int MAX_N = 200050;
20 const int MAX_LOG = 20;
21 typedef pair<int, int> pii;
22 int N, Q;
23 int p[MAX_N][MAX_LOG], depth[MAX_N];
24 vector<int> G[MAX_N];
25
26 void dfs(int u, int fa, int d) {
27 p[u][0] = fa;
28 depth[u] = d;
29 for (int i = 0; i < G[u].size(); i++) {
30 int v = G[u][i];
31 if (v != fa) dfs(v, u, d + 1);
32 }
33 }
34
35 void init() {
36 dfs(1, -1, 0);
37 for (int k = 0; k + 1 < MAX_LOG; k++) {
38 for (int v = 1; v <= N; v++) {
39 if (p[v][k] < 0) p[v][k + 1] = -1;
40 else p[v][k + 1] = p[p[v][k]][k];
41 }
42 }
43 }
44
45 int lca(int u, int v) {
46 if (depth[u] > depth[v]) swap(u, v);
47 for (int k = 0; k < MAX_LOG; k++) {
48 if ((depth[v] - depth[u]) >> k & 1) {
49 v = p[v][k];
50 }
51 }
52 if (u == v) return u;
53 for (int k = MAX_LOG - 1; k >= 0; k--) {
54 if (p[u][k] != p[v][k]) {
55 u = p[u][k];
56 v = p[v][k];
57 }
58 }
59 return p[u][0];
60 }
61
62 int dist(int u, int v) {
63 int x = lca(u, v);
64 return depth[u] + depth[v] - 2 * depth[x];
65 }
66
67 int main(void) {
68 ios::sync_with_stdio(false);
69 while (cin >> N) {
70 for (int i = 1; i <= N; i++) G[i].clear();
71 for (int i = 1; i < N; i++) {
72 int u, v;
73 cin >> u >> v;
74 G[u].push_back(v);
75 G[v].push_back(u);
76 }
77
78 init();
79 cin >> Q;
80 pii s[6];
81 while (Q--) {
82 int r, u, v;
83 cin >> r >> u >> v;
84 s[0].second = r;
85 s[1].second = u;
86 s[2].second = v;
87 s[3].second = lca(r, u);
88 s[4].second = lca(r, v);
89 s[5].second = lca(u, v);
90 for (int i = 0; i < 6; i++) {
91 int x = s[i].second;
92 s[i].first = dist(u, x) + dist(v, x) + dist(r, x);
93 }
94 sort(s, s + 6);
95 cout << s[0].second << endl;
96 }
97 }
98 return 0;
99 }
来源：https://www.cnblogs.com/Stomach-ache/p/3991434.html