[HEOI2014]大工程

description

题面

data range

[nle 10^6, qle 5 imes 10^4,sum kle 2 imes n ]

solution

还是虚树的题

这道题就相对比较容易了

直接建出虚树，考虑每条边对答案的贡献，再在树上做一个最长链和最短链就可以了

之前似乎一直在和空气斗智斗勇来着

code

#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<iomanip>
#include<cstring>
#include<complex>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<ctime>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define FILE "a"
#define mp make_pair
#define pb push_back
#define RG register
#define il inline
using namespace std;
typedef unsigned long long ull;
typedef vector<int>VI;
typedef long long ll;
typedef double dd;
const dd eps=1e-10;
const int mod=998244353;
const int N=2000010;
const dd pi=acos(-1);
const int inf=2147483647;
const ll INF=1e18+1;
const ll P=100000;
il ll read(){
    RG ll data=0,w=1;RG char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')w=-1,ch=getchar();
    while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
    return data*w;
}
 
il void file(){
    srand(time(NULL)+rand());
    freopen(FILE".in","r",stdin);
    freopen(FILE".out","w",stdout);
}
 
int n,q,m,k;
int head[N],nxt[N<<1],to[N<<1],cnt;
int dhead[N],dnxt[N<<1],dto[N<<1],dcnt;
il void addedge(int u,int v){
    dto[++dcnt]=v;
    dnxt[dcnt]=dhead[u];
    dhead[u]=dcnt;
}
 
int fa[N],sz[N],dep[N],son[N],top[N],dfn[N],low[N],tot;
void dfs1(int u,int ff){
    fa[u]=ff;dep[u]=dep[ff]+1;sz[u]=1;
    for(RG int i=head[u];i;i=nxt[i]){
        RG int v=to[i];if(v==ff)continue;
        dfs1(v,u);sz[u]+=sz[v];
        if(sz[son[u]]<sz[v])son[u]=v;
    }   
}
void dfs2(int u,int tp){
    top[u]=tp;dfn[u]=++tot;
    if(son[u])dfs2(son[u],tp);
    for(RG int i=head[u];i;i=nxt[i]){
        RG int v=to[i];if(v==fa[u]||v==son[u])continue;
        dfs2(v,v);
    }
    low[u]=++tot;
}
il 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]?u:v;
}
 
bool cmp_dfn(int i,int j){return dfn[i]<dfn[j];}
int s[N],cal[N],tp,mark[N],siz[N];
ll f[N],g[N],sum,maxn,minn;
void solve(int u){
    siz[u]=mark[u];
    for(RG int i=dhead[u];i;i=dnxt[i]){
        RG int v=dto[i];solve(v);siz[u]+=siz[v];
    }
    for(RG int i=dhead[u];i;i=dnxt[i]){
        RG int v=dto[i],d=dep[v]-dep[u];
        sum+=1ll*d*siz[v]*(m-siz[v]);
        maxn=max(maxn,f[u]+f[v]+d);
        minn=min(minn,g[u]+g[v]+d);
        f[u]=max(f[u],f[v]+d);
        g[u]=min(g[u],g[v]+d);
    }
    if(mark[u]){maxn=max(maxn,f[u]);if(g[u])minn=min(minn,g[u]);}
    if(mark[u])g[u]=0,f[u]=max(f[u],0ll);
}
 
int main()
{
    n=read();
    for(RG int i=1,u,v;i<n;i++){
        u=read();v=read();
        to[++cnt]=v;nxt[cnt]=head[u];head[u]=cnt;
        to[++cnt]=u;nxt[cnt]=head[v];head[v]=cnt;       
    }
    dfs1(1,0);dfs2(1,1);
    for(RG int i=1;i<=n;i++)f[i]=-INF,g[i]=INF;
     
    q=read();
    for(RG int i=1;i<=q;i++){
        m=read();k=tp=dcnt=0;
        for(RG int j=1;j<=m;j++)s[++k]=read(),mark[s[k]]=1;
        sort(s+1,s+k+1,cmp_dfn);
        for(RG int j=1;j<m;j++)s[++k]=lca(s[j],s[j+1]);s[++k]=1;
        sort(s+1,s+k+1,cmp_dfn);k=unique(s+1,s+k+1)-s-1;
        for(RG int j=1;j<=k;j++){
            while(tp&&low[cal[tp]]<dfn[s[j]])tp--;
            if(tp)addedge(cal[tp],s[j]);cal[++tp]=s[j];
        }
        sum=0;maxn=0;minn=INF;solve(1);
        printf("%lld %lld %lld
",sum,minn,maxn);
        for(RG int j=1;j<=k;j++)
            dhead[s[j]]=mark[s[j]]=0,f[s[j]]=-INF,g[s[j]]=INF;
    }
    return 0;
}