题面
虚树
把跟询问有关的点拿出来建树，为了方便树(DP)
在(LCA)处要合并答案，那么把这些点的(LCA)也拿出来
做法：把点按(dfs)序排列，然后求出相邻两个点的(LCA)，把这些点建一个虚树，维护一个栈就好了
Sol
虚树+树(DP)
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1e6 + 5);
const int inf(1e9);
int n, first1[maxn], cnt, first2[maxn], id[maxn], p[maxn];
ll g[maxn], f[maxn], num[maxn], sum, mx, mn, tot;
int dfn[maxn], size[maxn], son[maxn], fa[maxn];
int s[maxn], top[maxn], idx, deep[maxn];
struct Edge{
int to, next;
} e2[maxn << 1], e1[maxn << 1];
IL void Add1(RG int u, RG int v){
e1[cnt] = (Edge){v, first1[u]}, first1[u] = cnt++;
}
IL void Add2(RG int u, RG int v){
e2[cnt] = (Edge){v, first2[u]}, first2[u] = cnt++;
}
IL void Dfs1(RG int u){
size[u] = 1;
for(RG int e = first1[u]; e != -1; e = e1[e].next){
RG int v = e1[e].to;
if(!size[v]){
deep[v] = deep[u] + 1;
Dfs1(v), size[u] += size[v], fa[v] = u;
if(size[v] > size[son[u]]) son[u] = v;
}
}
}
IL void Dfs2(RG int u, RG int tp){
top[u] = tp, dfn[u] = ++idx;
if(son[u]) Dfs2(son[u], tp);
for(RG int e = first1[u]; e != -1; e = e1[e].next)
if(!dfn[e1[e].to]) Dfs2(e1[e].to, e1[e].to);
}
IL int LCA(RG int u, RG int v){
while(top[u] ^ top[v])
deep[top[u]] > deep[top[v]] ? u = fa[top[u]] : v = fa[top[v]];
return deep[u] > deep[v] ? v : u;
}
IL int Dis(RG int u, RG int v){
RG int lca = LCA(u, v);
return deep[u] + deep[v] - 2 * deep[lca];
}
IL int Cmp(RG int u, RG int v){
return dfn[u] < dfn[v];
}
IL void DP(RG int u){
g[u] = inf, f[u] = -inf;
if(num[u]) g[u] = f[u] = 0;
for(RG int e = first2[u]; e != -1; e = e2[e].next){
RG int v = e2[e].to, w = Dis(u, v);
DP(v), num[u] += num[v];
sum += (tot - num[v]) * num[v] * w;
mn = min(mn, g[u] + w + g[v]);
mx = max(mx, f[u] + w + f[v]);
g[u] = min(g[u], g[v] + w);
f[u] = max(f[u], f[v] + w);
}
}
int main(){
n = Input();
for(RG int i = 1; i <= n; ++i) first1[i] = first2[i] = -1;
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
Add1(u, v), Add1(v, u);
}
Dfs1(1), Dfs2(1, 1);
for(RG int q = Input(); q; --q){
RG int k = Input(); cnt = 0, tot = k;
for(RG int i = 1; i <= k; ++i) p[i] = Input(), num[p[i]] = 1;
sort(p + 1, p + k + 1, Cmp);
for(RG int i = 1, t = k; i < t; ++i) p[++k] = LCA(p[i], p[i + 1]);
sort(p + 1, p + k + 1, Cmp), k = unique(p + 1, p + k + 1) - p - 1;
for(RG int i = 1; i <= k; ++i) first2[p[i]] = -1;
RG int t = 0;
for(RG int i = 1; i <= k; ++i){
while(t && dfn[p[i]] >= dfn[s[t]] + size[s[t]]) --t;
if(t) Add2(s[t], p[i]);
s[++t] = p[i];
}
mx = sum = 0, mn = inf;
DP(p[1]);
printf("%lld %lld %lld
", sum, mn, mx);
for(RG int i = 1; i <= k; ++i) num[p[i]] = 0;
}
return 0;
}