傳送門：http://www.lydsy.com/JudgeOnline/problem.php?id=3611

【題解】

和bzoj2286一樣，建虛樹出來，在虛數上dp。

第一問分別考慮每條邊的貢獻即可。

第二問和第三問dp。

f[x][0/1],g[x][0/1]表示x子樹，兩個端點都在子樹內/一個端點在子樹內的max/min

轉移隨便討論討論

只是有點兒麻煩。。

第一問沒longlong。。WA了好久qwq

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
//
 
 # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 2e6 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n;
struct Graph {
    int n, head[M], nxt[M], to[M], w[M], tot;
    inline 
 
void set(int _n) {
        n = _n;
        tot = 0;
        for (int i=1; i<=n; ++i) head[i] = 0;
    }
    inline void add(int u, int v, int _w) {
        ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; w[tot] = _w;
    }
    inline void adde(int u, int v, int _w=0) {
        add(u, v, _w);
        add(v, u, _w);
    }
}G, T;

 
int fa[M][21], dep[M], dfn[M], DFN=0;
inline void dfs(int x, int father=0) {
    dfn[x] = ++DFN;
    for (int i=1; i<=20; ++i)
        fa[x][i] = fa[fa[x][i-1]][i-1];
    for (int i=G.head[x]; i; i=G.nxt[i]) {
        if(G.to[i] == father) continue;
        fa[G.to[i]][0] = x;
        dep[G.to[i]] = dep[x] + 1;
        dfs(G.to[i], x);
    }
}

inline int lca(int u, int v) {
    if(dep[u] < dep[v]) swap(u, v);
    for (int i=20; ~i; --i) 
        if((dep[u] - dep[v]) & (1<<i)) u = fa[u][i];
    if(u == v) return u;
    for (int i=20; ~i; --i)
        if(fa[u][i] != fa[v][i]) {
            u = fa[u][i];
            v = fa[v][i];
        }
    return fa[u][0];
}

inline int getdis(int u, int v) {
    int d = 0;
    if(dep[u] < dep[v]) swap(u, v);
    for (int i=20; ~i; --i) 
        if((dep[u] - dep[v]) & (1<<i)) {
            u = fa[u][i];
            d += (1<<i);
        }
    if(u == v) return d;
    for (int i=20; ~i; --i)
        if(fa[u][i] != fa[v][i]) {
            u = fa[u][i];
            v = fa[v][i];
            d += (1<<i)*2;
        }
    return d + 2;
}

inline void Tadd(int u, int v) {
//    printf("%d->%d\n", u, v);
    T.adde(u, v, getdis(u, v));
}

ll ans = 0;
int sz[M];
bool g[M];
int m, a[M], st[M], stn;

inline void predp(int x, int father=0) {
    int wfa = 0;
    sz[x] = g[x];
    for (int i=T.head[x]; i; i=T.nxt[i]) {
        int y = T.to[i];
        if(y == father) {
            wfa = T.w[i];
            continue;
        }
        predp(y, x);
        sz[x] += sz[y];
    }
    if(father != 0) {
        // x->father
        ans += 1ll * wfa * sz[x] * (m-sz[x]);
    }
}

//f: min, h:max
// first: all-in, second: one-out
pair<int, int> f[M], h[M];
inline void dp(int x, int father=0) {
    f[x] = make_pair(1e9, 1e9);
    h[x] = make_pair(-1, -1);
    int ffir = 1e9, fsec = 1e9, hfir = -1, hsec = -1;
    for (int i=T.head[x], tem; i; i=T.nxt[i]) {
        int y = T.to[i];
        if(y == father) continue;
        dp(y, x);
        if(f[y].second != 1e9) {
            tem = f[y].second + T.w[i];
            f[x].second = min(f[x].second, tem);
            if(tem < ffir) fsec = ffir, ffir = tem;
            else if(tem < fsec) fsec = tem;
        }
        f[x].first = min(f[x].first, f[y].first);
        if(h[y].second != -1) {
            tem = h[y].second + T.w[i];
            h[x].second = max(h[x].second, tem);
            if(tem > hfir) hsec = hfir, hfir = tem;
            else if(tem > hsec) hsec = tem;
        }
        h[x].first = max(h[x].first, h[y].first);
    }
    if(g[x]) {
        // is a dot
        f[x].second = min(f[x].second, 0);
        h[x].second = max(h[x].second, 0);
        f[x].first = min(f[x].first, ffir);
        if(hfir != -1) {
            if(hsec == -1) h[x].first = max(h[x].first, hfir);
            else h[x].first = max(h[x].first, hfir + hsec);
        }
    } else {
        f[x].first = min(f[x].first, ffir + fsec);
        if(hfir != -1 && hsec != -1) h[x].first = max(h[x].first, hfir + hsec);
    }
    T.head[x] = 0;
}

inline bool cmp_dfn(int x, int y) {
    return dfn[x] < dfn[y];
}

inline void sol() {
//    puts("====");
    ans = 0;
    scanf("%d", &m);
    for (int i=1; i<=m; ++i) scanf("%d", &a[i]);
    if(m == 0 || m == 1) {
        puts("0 0 0");
        return ;
    }
    sort(a+1, a+m+1, cmp_dfn);
    for (int i=1; i<=m; ++i) g[a[i]] = 1;
    stn = 0; T.tot = 0;
    st[++stn] = 1;
    for (int i=1; i<=m; ++i) {
        if(a[i] == 1) continue;
        int LCA = lca(st[stn], a[i]);
        while(stn && dep[st[stn]] > dep[LCA]) {
            if(stn > 1) {
                if(dep[st[stn-1]] > dep[LCA]) Tadd(st[stn-1], st[stn]);
                else Tadd(st[stn], LCA);
            }
            --stn;
        }
        if(st[stn] != LCA) st[++stn] = LCA;
        st[++stn] = a[i];
    }
    while(stn > 1) {
        Tadd(st[stn], st[stn-1]);
        --stn;
    }
    predp(1);
    dp(1);
    printf("%lld %d %d\n", ans, f[1].first, h[1].first);
    for (int i=1; i<=m; ++i) g[a[i]] = 0;
}

int main() {
    cin >> n;
    for (int i=1, u, v; i<n; ++i) {
        scanf("%d%d", &u, &v);
        G.adde(u, v);
    } 
    dep[1] = 1;
    dfs(1);
    int Q; cin >> Q;
    while(Q--) sol();
    return 0;
}

bzoj3611 [Heoi2014]大工程