Description

题库链接

给你一个 \(n\) 个节点的树， \(q\) 组询问，每次给出 \(k\) 个关键点，询问这 \(k\) 个关键点两两间路径长度和，长度最值。

\(1\leq n\leq 1000000, 1\leq q\leq 50000,\sum k\leq 2n\)

Solution

建完虚树后，直接简单的树形 \(DP\) ，开四个辅助数组， \(size,sum,maxn,minn\) 表示虚树中以它为根的子树中所有关键点的个数，到它的路径长度和，长度最值。

再一顿乱搞就好了。

Code

//It is made by Awson on 2018.2.21

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 1000000;

const int INF = ~0u>>1;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, lim, fa[N+5][25], dep[N+5], dfn[N+5], times, u, v;

LL maxans, minans, sumans, minn[N+5], maxn[N+5], sum[N+5], size[N+5];

int lst[N+5], flag[N+5], k, t, S[N+5], top;

struct graph {

    struct tt {int to, next; }edge[(N<<1)+5];

    int path[N+5], top;

    void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }

    void dfs1(int o, int depth) {

    dep[o] = depth, dfn[o] = ++times;

    for (int i = 1; i <= lim; i++) fa[o][i] = fa[fa[o][i-1]][i-1];

    for (int i = path[o]; i; i = edge[i].next)

        if (dfn[edge[i].to] == 0) fa[edge[i].to][0] = o, dfs1(edge[i].to, depth+1);

    }

    void dfs2(int o) {

    minn[o] = INF, maxn[o] = -INF, sum[o] = 0, size[o] = 0;

    if (flag[o]) minn[o] = maxn[o] = 0, size[o] = 1;

    for (int &i = path[o]; i; i = edge[i].next) {

        dfs2(edge[i].to);

        sumans += size[o]*(sum[edge[i].to]+(dep[edge[i].to]-dep[o])*size[edge[i].to])+sum[o]*size[edge[i].to];

        sum[o] += sum[edge[i].to]+size[edge[i].to]*(dep[edge[i].to]-dep[o]), size[o] += size[edge[i].to];

        minans = Min(minans, minn[o]+minn[edge[i].to]+dep[edge[i].to]-dep[o]), minn[o] = Min(minn[o], minn[edge[i].to]+dep[edge[i].to]-dep[o]);

        maxans = Max(maxans, maxn[o]+maxn[edge[i].to]+dep[edge[i].to]-dep[o]), maxn[o] = Max(maxn[o], maxn[edge[i].to]+dep[edge[i].to]-dep[o]);

    }

    }

}g1, g2;

bool comp(const int &a, const int &b) {return dfn[a] < dfn[b]; }

int get_lca(int u, int v) {

    if (dep[u] < dep[v]) Swap(u, v);

    for (int i = lim; i >= 0; i--) if (dep[fa[u][i]] >= dep[v]) u = fa[u][i];

    if (u == v) return u;

    for (int i = lim; i >= 0; i--) if (fa[u][i] != fa[v][i]) u = fa[u][i], v = fa[v][i];

    return fa[u][0];

}

void work() {

    read(n); lim = log(n)/log(2);

    for (int i = 1; i < n; i++) read(u), read(v), g1.add(u, v), g1.add(v, u);

    g1.dfs1(1, 1); read(t);

    while (t--) {

    read(k); top = g2.top = 0; maxans = -INF, minans = INF, sumans = 0;

    for (int i = 1; i <= k; i++) read(lst[i]), flag[lst[i]] = 1;

    sort(lst+1, lst+1+k, comp); S[++top] = lst[1];

    for (int i = 2; i <= k; i++) {

        int lca = get_lca(lst[i], S[top]);

        while (dfn[lca] < dfn[S[top]]) {

        if (dfn[lca] >= dfn[S[top-1]]) {

            g2.add(lca, S[top]); --top;

            if (S[top] != lca) S[++top] = lca;

            break;

        }

        g2.add(S[top-1], S[top]), --top;

        }

        S[++top] = lst[i];

    }

    while (top > 1) g2.add(S[top-1], S[top]), --top;

    g2.dfs2(S[1]);

    write(sumans), putchar(' '), write(minans), putchar(' '), writeln(maxans);

    for (int i = 1; i <= k; i++) flag[lst[i]] = 0;

    }

}

int main() {

    work(); return 0;

}