题意:
题解:
AC代码
/*
Author : zzugzx
Lang : C++
Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
//const int mod = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 1e2 + 5;
const ll inf = 0x3f3f3f3f;
const ll oo = 8e18;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
int n,depth[maxn], f[maxn][50];
int from[maxn], to[maxn << 1], nxt[maxn << 1], cnt,Log[maxn];
//链式前向星加边
void addEdge (int u, int v) {
to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}
//计算深度&计算祖先
void dfs (int u, int fa) {
depth[u] = depth[fa] + 1;
for (register int i = 1; i <= Log[n]; ++i) {
if ((1 << i) > depth[u]) break;
f[u][i] = f[f[u][i - 1]][i - 1];
}
for (register int i = from[u]; i; i = nxt[i]) {
ll v = to[i];
if (v == fa) continue;
f[v][0] = u;
dfs (v, u);
}
}
//计算LCA
inline int LCA (int x, int y) {
if (depth[x] < depth[y]) swap(x, y);
//我们默认x为更深的那个点
for(register int i = Log[n] ; i >= 0 ; --i)
if(depth[x] - (1 << i) >= depth[y]) x = f[x][i];
//将x跳到和y同一深度上
if (x == y) return x;
for (register int i = Log[n]; i >= 0; --i)
if (f[x][i] != f[y][i])
x = f[x][i], y = f[y][i];
//一起向上跳
return f[x][0];
//不难看出,此时两个点均在其LCA的下方,往上跳一次即可
}
void init(){
Log[0] = -1;
cnt = 0;
for(int i = 1; i <= n; i++){
from[i] = 0,depth[i] = 0;
for(int j = 0; j < 50; j++)
f[i][j] = 0;
}
for (register int i = 1, u, v; i < n; ++i) {
cin >> u >> v;
addEdge (u, v); addEdge(v, u);
Log[i] = Log[i >> 1] + 1;
}
Log[n] = Log[n >> 1] + 1;
dfs(1,0);
}
int dis(int p , int q){return depth[p] + depth[q] - 2 * depth[LCA(p , q)];}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
cin >> n;
init();
int _;
cin >> _;
while (_--) {
int k, x, y = 0, ans = 0;
cin >> k;
vector<int> v;
for(int i = 1; i <= k; i++) {
cin >> x;
if (depth[x] > depth[y]) y = x;
v.pb(x);
}
for (auto i : v) ans = max(ans, dis(y, i));
cout << (ans + 1) /2 << endl;
}
return 0;
}

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