Lowest Common Ancestor (LCA)

题目链接

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
2

Explanation

• "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:

 /*************************************************************************
     > File Name: TALCA.cpp
     > Author: Stomach_ache
     > Mail: sudaweitong@gmail.com
     > Created Time: 2014年09月24日 星期三 17时39分16秒
     > Propose:
  ************************************************************************/
 #include <cmath>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <fstream>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 /*Let's fight!!!*/

 const int MAX_N = ;
 const int MAX_LOG = ;
 typedef pair<int, int> pii;
 int N, Q;
 int p[MAX_N][MAX_LOG], depth[MAX_N];
 vector<int> G[MAX_N];

 void dfs(int u, int fa, int d) {
     p[u][] = fa;
     depth[u] = d;
     for (int i = ; i < G[u].size(); i++) {
         int v = G[u][i];
         if (v != fa) dfs(v, u, d + );
     }
 }

 void init() {
     dfs(, -, );
     for (int k = ; k +  < MAX_LOG; k++) {
           for (int v = ; v <= N; v++) {
             if (p[v][k] < ) p[v][k + ] = -;
             else p[v][k + ] = p[p[v][k]][k];
         }
     }
 }

 int lca(int u, int v) {
     if (depth[u] > depth[v]) swap(u, v);
     for (int k = ; k < MAX_LOG; k++) {
         if ((depth[v] - depth[u]) >> k & ) {
               v = p[v][k];
         }
     }
     if (u == v) return u;
     for (int k = MAX_LOG - ; k >= ; k--) {
           if (p[u][k] != p[v][k]) {
             u = p[u][k];
             v = p[v][k];
         }
     }
     return p[u][];
 }

 int dist(int u, int v) {
       int x = lca(u, v);
       return depth[u] + depth[v] -  * depth[x];
 }

 int main(void) {
     ios::sync_with_stdio(false);
     while (cin >> N) {
         for (int i = ; i <= N; i++) G[i].clear();
         for (int i = ; i < N; i++) {
             int u, v;
             cin >> u >> v;
             G[u].push_back(v);
             G[v].push_back(u);
         }

         init();
         cin >> Q;
         pii s[];
         while (Q--) {
             int r, u, v;
             cin >> r >> u >> v;
             s[].second = r;
             s[].second = u;
             s[].second = v;
             s[].second = lca(r, u);
             s[].second = lca(r, v);
             s[].second = lca(u, v);
             for (int i = ; i < ; i++) {
                   int x = s[i].second;
                   s[i].first = dist(u, x) + dist(v, x) + dist(r, x);
             }
             sort(s, s + );
             cout << s[].second << endl;
         }
     }
     return ;
 }