[HNOI 2014]世界树

Description

题库链接

给出一棵 \(n\) 个节点的树， \(q\) 次询问，每次给出 \(k\) 个关键点。树上所有的点会被最靠近的关键点管辖，若距离相等则选编号最小的那个。求每个关键点管辖多少个节点。

\(1\leq n,q,\sum k\leq 300000\)

Solution

构出虚树后，我们能用简单的树形 \(dp\) 求出每个点离他最近的关键点。大体是做两遍 \(dfs\) 。第一遍用儿子更新父亲，第二遍用父亲更新儿子。

处理好这个之后，对于虚树上每个点。他的子树有两种：一个是虚树里的，一个是不在虚树里的。不在虚树里的后代肯定和他共用同一个关键点；

对于虚树上的一条边 \((u,v)\) ，我们需要找到 \((u,v)\) 边上的所有点以及他们连出去的块的最近点，更新答案。

如果 \(u, v\) 的最近点相同，那么这条边所代表的所有点的最近点肯定就是 \(u,v\) 的最近点；否则，可以用倍增找到临界点，计算贡献。

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 = 300000;
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(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, lim, u, v, fa[N+5][20], dep[N+5], size[N+5], dfn[N+5], times;
int flag[N+5], lst[N+5], kp[N+5], belong[N+5], dist[N+5], k, q, ans[N+5], 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, size[o] = 1, 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), size[o] += size[edge[i].to];
    }
    void dfs2(int o) {
    belong[o] = 0, dist[o] = INF>>1;
    if (flag[o]) belong[o] = o, dist[o] = 0;
    for (int i = path[o]; i; i = edge[i].next) {
        dfs2(edge[i].to);
        if ((dist[edge[i].to]+dep[edge[i].to]-dep[o] < dist[o]) || (dist[edge[i].to]+dep[edge[i].to]-dep[o] == dist[o] && belong[o] > belong[edge[i].to])) dist[o] = dist[edge[i].to]+dep[edge[i].to]-dep[o], belong[o] = belong[edge[i].to];
    }
    }
    void dfs3(int o) {
    for (int i = path[o]; i; i = edge[i].next) {
        if ((dist[o]+dep[edge[i].to]-dep[o] < dist[edge[i].to]) || (dist[o]+dep[edge[i].to]-dep[o] == dist[edge[i].to] && belong[o] < belong[edge[i].to])) dist[edge[i].to] = dist[o]+dep[edge[i].to]-dep[o], belong[edge[i].to] = belong[o];
        dfs3(edge[i].to);
    }
    }
    void dfs4(int o) {
    int rem = size[o];
    for (int &i = path[o]; i; i = edge[i].next) {
        int x = edge[i].to; for (int j = lim; j >= 0; j--) if (dep[fa[x][j]] > dep[o]) x = fa[x][j];
        rem -= size[x];
        if (belong[edge[i].to] == belong[o]) ans[belong[o]] += size[x]-size[edge[i].to];
        else {
        int v = edge[i].to;
        for (int j = lim; j >= 0; j--)
            if (dep[fa[v][j]] >= dep[o])
            if ((dist[edge[i].to]+dep[edge[i].to]-dep[fa[v][j]] < dist[o]+dep[fa[v][j]]-dep[o]) || (dist[edge[i].to]+dep[edge[i].to]-dep[fa[v][j]] == dist[o]+dep[fa[v][j]]-dep[o] && belong[edge[i].to] < belong[o])) v = fa[v][j];
        ans[belong[o]] += size[x]-size[v];
        ans[belong[edge[i].to]] += size[v]-size[edge[i].to];
        }
        dfs4(edge[i].to);
    }
    ans[belong[o]] += rem;
    }
}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(q);
    while (q--) {
    read(k); for (int i = 1; i <= k; i++) read(lst[i]), flag[kp[i] = lst[i]] = 1;
    top = g2.top = 0; sort(lst+1, lst+1+k, comp);
    S[++top] = 1;
    for (int i = 1; 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;
        }
        if (S[top] != lst[i]) S[++top] = lst[i];
    }
    while (top > 1) g2.add(S[top-1], S[top]), --top;
    g2.dfs2(1), g2.dfs3(1), g2.dfs4(1);
    for (int i = 1; i < k; i++) write(ans[kp[i]]), putchar(' ');
    writeln(ans[kp[k]]);
    for (int i = 1; i <= k; i++) flag[kp[i]] = ans[kp[i]] = 0;
    }
}
int main() {
    work(); return 0;
}

原文地址：https://www.cnblogs.com/NaVi-Awson/p/8457120.html