传送门

首先肯定建出虚树
考虑三种答案如何分别统计

路径长度和显然示是对于每条边考虑一下上下有多少个点，乘一下就可以了
最大值显然就是树的直径
最小值可以考虑树形dpdpdp，考虑mn[i]mn[i]mn[i]表示iii到其子树中最近的距离
显然可以类似直径树形dpdpdp解决

#include<bits/stdc++.h>
using namespace std;
#define ll long long
inline int read(){char ch=getchar();int res=0,f=1;while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=getchar();return res*f;
}
const int N=1000005;
const ll inf=1e9;
namespace T{int adj[N],nxt[N<<1],to[N<<1],cnt,siz[N],dfn[N],tot,son[N],top[N],fa[N],dep[N];inline void addedge(int u,int v){nxt[++cnt]=adj[u],adj[u]=cnt,to[cnt]=v;}void dfs1(int u){siz[u]=1;for(int e=adj[u];e;e=nxt[e]){int v=to[e];if(v==fa[u])continue;fa[v]=u,dep[v]=dep[u]+1;dfs1(v),siz[u]+=siz[v];if(siz[v]>siz[son[u]])son[u]=v;}}void dfs2(int u,int tp){dfn[u]=++tot,top[u]=tp;if(son[u])dfs2(son[u],tp);for(int e=adj[u];e;e=nxt[e]){int v=to[e];if(v==fa[u]||v==son[u])continue;dfs2(v,v);}}inline int Lca(int u,int v){while(top[u]!=top[v]){if(dep[top[u]]<dep[top[v]])swap(u,v);u=fa[top[u]];}return dep[u]>dep[v]?v:u;}
};
int n,adj[N],nxt[N<<1],to[N<<1],cnt,q,a[N],stk[N],top,k,vis[N],siz[N],ans2,ans3,mx[N],mn[N];
ll ans1;
inline void addedge(int u,int v){nxt[++cnt]=adj[u],adj[u]=cnt,to[cnt]=v;
}
inline bool comp(const int &a,const int &b){return T::dfn[a]<T::dfn[b];
}
inline void chemn(int &a,int b){a=a>b?b:a;
}
inline void chemx(int &a,int b){a=a>b?a:b;
}
void dp(int u){mx[u]=0,mn[u]=inf,siz[u]=vis[u];for(int e=adj[u];e;e=nxt[e]){int v=to[e];int d=T::dep[v]-T::dep[u];dp(v),siz[u]+=siz[v];ans1+=1ll*siz[v]*(k-siz[v])*d;chemn(ans2,mn[u]+mn[v]+d),chemn(mn[u],mn[v]+d);chemx(ans3,mx[u]+mx[v]+d),chemx(mx[u],mx[v]+d);}if(vis[u])chemn(ans2,mn[u]),chemx(ans3,mx[u]),mn[u]=0;adj[u]=0;
}
inline void work(){top=0,cnt=0;for(int i=1;i<=k;i++){if(!top){stk[++top]=a[i];continue;}int lca=T::Lca(stk[top],a[i]);while(T::dfn[lca]<T::dfn[stk[top]]){if(T::dfn[lca]>=T::dfn[stk[top-1]]){addedge(lca,stk[top]);if(stk[--top]!=lca)stk[++top]=lca;break;}addedge(stk[top-1],stk[top]),top--;}stk[++top]=a[i];}while(top>1)addedge(stk[top-1],stk[top]),top--;ans1=0,ans2=inf,ans3=0,dp(stk[1]);for(int i=1;i<=k;i++)vis[a[i]]=0;cout<<ans1<<" "<<ans2<<" "<<ans3<<'\n';
}
int main(){n=read();for(int i=1;i<n;i++){int u=read(),v=read();T::addedge(u,v),T::addedge(v,u);}T::dfs1(1),T::dfs2(1,1);q=read();for(int i=1;i<=q;i++){k=read();for(int j=1;j<=k;j++)a[j]=read(),vis[a[j]]=1;sort(a+1,a+k+1,comp),work();}
}