Lowest Common Ancestor（CodeChef TALCA）

In a rooted tree, the lowest common ancestor (or LCA for short) of two vertices u andv 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 atr 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 integeru and v 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 nextQ 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:41 22 31 421 4 22 4 2Output:12

 

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.

https://vjudge.net/problem/CodeChef-TALCA

给一棵有n个节点的树，n-1条双向边
先给q个询问，
每个询问格式； root u v
    求以root为根的树，u与v的最近公共祖先
    
DFS+并查集的LCA问题

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <sstream>#include <map>#include <set>#define pi acos(-1.0)#define LL long long#define ULL unsigned long long#define inf 0x3f3f3f3f#define INF 1e18#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const double eps = 1e-10;const int maxn = 2e5 + 5;const int N = 1e4 + 5;const int mod = 1e8;int par[maxn];int vis[maxn];int n, q, flag;int root, node1, node2;vector<int>G[maxn]; int find(int a){if (par[a] == -1) return a;return par[a] = find(par[a]);}void unite(int a, int b){int x = find(a);int y = find(b);par[y] = x;}void LCA(int u){vis[u] = 1;for (int i = 0; i < G[u].size(); i++){int v = G[u][i];if (vis[v]) continue;LCA(v);unite(u, v);}if (flag) return; // 防止输出两次 if (u == node1 && vis[node2]){cout << find(node2) << endl;flag = 1;return;}if (u == node2 && vis[node1]){cout << find(node1) << endl;flag = 1;return;}}int main(void){//freopen("in.txt", "r", stdin);cin >> n;int u, v;for (int i = 1; i < n; i++){cin >> u >> v;G[u].push_back(v);G[v].push_back(u);}cin >> q;for (int i = 1; i <= q; i++){cin >> root >> node1 >> node2; memset(vis, 0, sizeof(vis));memset(par, -1, sizeof(par));flag = 0;LCA(root);}return 0;}