题意
Sol
虚树。
首先建出虚树，然后直接树形dp就行了。
最大最小值直接维护子树内到该节点的最大值，然后合并两棵子树的时候更新一下答案。
任意两点的路径和可以考虑每条边两边的贡献，(d[x])表示到该节点的所有节点的路径和，转移的时候考虑一下两棵子树的siz就行(画一下图就很清楚了)
每次询问可以用个队列记录一下被访问过的元素，清空的时候只清空队列里的点。
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long
#define LL long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 2e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10;
const double eps = 1e-9;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename A> inline void debug(A a){cout << a << '
';}
template <typename A> inline LL sqr(A x){return 1ll * x * x;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, dfn[MAXN];
vector<int> v[MAXN], E[MAXN];
namespace HLP {
int fa[MAXN], siz[MAXN], son[MAXN], dep[MAXN], top[MAXN], times;
void dfs1(int x, int _fa) {
siz[x] = 1; fa[x] = _fa; dep[x] = dep[_fa] + 1; dfn[x] = ++times;
for(auto &to : v[x]) {
if(to == _fa) continue;
dfs1(to, x);
siz[x] += siz[to];
if(siz[to] > siz[son[x]]) son[x] = to;
}
}
void dfs2(int x, int topf) {
top[x] = topf;
if(!son[x]) return ;
dfs2(son[x], topf);
for(auto &to : v[x]) {
if(top[to]) continue;
dfs2(to, to);
}
}
int LCA(int x, int y) {
while(top[x] ^ top[y]) {
if(dep[top[x]] < dep[top[y]]) swap(x, y);
x = fa[top[x]];
}
return dep[x] < dep[y] ? x : y;
}
}
int p[MAXN], st[MAXN], top, f[MAXN], mx[MAXN], mn[MAXN], siz2[MAXN], d[MAXN], ans1, ans2, ans3;
void AE(int x, int y) {E[x].push_back(y);}
int comp(int a, int b) {return dfn[a] < dfn[b];}
int dis(int x, int y) {if(dfn[x] > dfn[y]) swap(x, y); return HLP::dep[y] - HLP::dep[x];}
queue<int> q;
void insert(int x) {
if(top == 1) {st[++top] = x; return ;}
int lca = HLP::LCA(x, st[top]);
if(lca == st[top]) {st[++top] = x; return ;}
while(top > 1 && dfn[st[top - 1]] >= dfn[lca]) AE(st[top - 1], st[top]), top--;
if(lca != st[top])
AE(lca, st[top]), st[top] = lca;
st[++top] = x;
}
//ans1 和 ans2最大值 ans3最小值
void DP(int x) {
q.push(x);
mn[x] = (siz2[x] ? 0 : 1e18);
mx[x] = 0; d[x] = 0;
for(auto &to : E[x]) {
int w = dis(x, to);
assert(w <= N);
DP(to);
if(siz2[x] > 0) {
ans1 += w * siz2[to] * siz2[x] + siz2[to] * d[x] + siz2[x] * d[to];
chmax(ans2, mx[x] + mx[to] + w);
chmin(ans3, mn[x] + mn[to] + w);
}
chmax(mx[x], w + mx[to]);
chmin(mn[x], w + mn[to]);
siz2[x] += siz2[to];
d[x] += d[to] + w * siz2[to];
}
E[x].clear();
}
void work() {
ans1 = ans2 = 0; ans3 = INF;
int K = read(); st[top = 1] = 1;
for(int i = 1; i <= K; i++) p[i] = read();
sort(p + 1, p + K + 1, comp);
for(int i = 1; i <= K; i++) {
if(p[i] != 1) insert(p[i]);
siz2[p[i]] = 1;
}
while(top > 1) AE(st[top - 1], st[top]), top--;
DP(1);
while(!q.empty()) siz2[q.front()] = 0, q.pop();
cout << ans1 << ' ' << ans3 << ' ' << ans2 << '
';
}
signed main() {
N = read();
for(int i = 1; i <= N - 1; i++) {
int x = read(), y = read();
v[x].push_back(y);
v[y].push_back(x);
}
HLP::dfs1(1, 0); HLP::dfs2(1, 1);
int Q = read();
while(Q--) {
work();
}
return 0;
}