题面

传送门

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\n", sum, mn, mx);
        for(RG int i = 1; i <= k; ++i) num[p[i]] = 0;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/cjoieryl/p/9113969.html