- 这个题貌似是过的最少的?
- smeow一眼给出了一个单log的算法orz
- 首先求出x和y的lca, x和c的lca,y和c的lca, 然后分类讨论以下就行了
- 实际上只有三种情况
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 100010
#define mmp make_pair
using namespace std;
int read() {
int nm = 0, f = 1;
char c = getchar();
for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
return nm * f;
}
int fa[M], top[M], son[M], sz[M], n, q, deep[M];
vector<int> to[M];
void dfs(int now, int f) {
fa[now] = f;
sz[now] = 1;
deep[now] = deep[f] + 1;
for(int i = 0; i < to[now].size(); i++) {
int vj = to[now][i];
if(vj == f) continue;
dfs(vj, now);
if(sz[son[now]] < sz[vj]) son[now] = vj;
sz[now] += sz[vj];
}
}
void dfs(int now) {
if(son[now]) {
top[son[now]] = top[now];
dfs(son[now]);
}
for(int i = 0; i < to[now].size(); i++) {
int vj = to[now][i];
if(vj == fa[now] || vj == son[now]) continue;
top[vj] = vj;
dfs(vj);
}
}
int lca(int a, int b) {
for(; top[a] != top[b]; a = fa[top[a]]) {
if(deep[top[a]] < deep[top[b]]) swap(a, b);
}
if(deep[a] > deep[b]) swap(a, b);
return a;
}
int main() {
n = read();
for(int i = 1; i < n; i++) {
int vi = read(), vj = read();
to[vi].push_back(vj), to[vj].push_back(vi);
}
dfs(1, 0);
top[1] = 1;
dfs(1);
q = read();
while(q--) {
int x = read(), y = read(), c = read();
int l1 = lca(x, y), l2 = lca(x, c), l3 = lca(y, c), ans;
if(l2 == l1) ans = l3;
else if(l3 == l1) ans = l2;
else ans = l1;
cout << ans << "\n";
}
return 0;
}