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 = 200050;
const int MAX_LOG = 20;
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][0] = fa;
    depth[u] = d;
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (v != fa) dfs(v, u, d + 1);
    }
}

void init() {
    dfs(1, -1, 0);
    for (int k = 0; k + 1 < MAX_LOG; k++) {
          for (int v = 1; v <= N; v++) {
            if (p[v][k] < 0) p[v][k + 1] = -1;
            else p[v][k + 1] = p[p[v][k]][k];
        }
    }
}

int lca(int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    for (int k = 0; k < MAX_LOG; k++) {
        if ((depth[v] - depth[u]) >> k & 1) {
              v = p[v][k];
        }
    }
    if (u == v) return u;
    for (int k = MAX_LOG - 1; k >= 0; k--) {
          if (p[u][k] != p[v][k]) {
            u = p[u][k];
            v = p[v][k];
        }
    }
    return p[u][0];
}

int dist(int u, int v) {
      int x = lca(u, v);
      return depth[u] + depth[v] - 2 * depth[x];
}

int main(void) {
    ios::sync_with_stdio(false);
    while (cin >> N) {
        for (int i = 1; i <= N; i++) G[i].clear();
        for (int i = 1; i < N; i++) {
            int u, v;
            cin >> u >> v;
            G[u].push_back(v);
            G[v].push_back(u);
        }

        init();
        cin >> Q;
        pii s[6];
        while (Q--) {
            int r, u, v;
            cin >> r >> u >> v;
            s[0].second = r;
            s[1].second = u;
            s[2].second = v;
            s[3].second = lca(r, u);
            s[4].second = lca(r, v);
            s[5].second = lca(u, v);
            for (int i = 0; i < 6; i++) {
                  int x = s[i].second;
                  s[i].first = dist(u, x) + dist(v, x) + dist(r, x);
            }
            sort(s, s + 6);
            cout << s[0].second << endl;
        }
    }
    return 0;
}