题解CF#983 E-NN country

Posted 2021-01-17 twilight-sx

　　首先，我们从 u -> v 有一个明显的贪心，即能向上跳的时候尽量向深度最浅的节点跳。这个我们可以用树上倍增来维护。我们可以认为 u 贪心向上跳后不超过 lca 能跳到 u‘ 的位置， v 跳到 v‘ 的位置，这时只需要查询一下是否有 u‘ -> v‘ 的直达公交线路就可以确定出答案了。

　　如果 u 和 v 在一条链上，我们可以直接倍增获得答案，所以以下讨论均为 u != lca && v != lca 的情况。 若在节点 u 和 v 之间存在一条直达线路，说明有一条线路的起点在以 u 为根的子树内，重点在以 v 为根的子树内。即起点的dfs序 >= dfn[u] && <= dfn[u] + size[u] - 1, 终点的 dfs序 >= dfn[v] && <= dfn[v] + size[v] -1;这是一个二维限制的问题，暴力主席树就可以搞定……

#include <bits/stdc++.h>
using namespace std;
#define maxn 450000
#define INF 99999999
#define CNST 20
int n, m, q, Mul[maxn][CNST], gra[maxn][CNST];
int dfn[maxn], dep[maxn], bits[30];
int tot, timer, root[maxn], fa[maxn], size[maxn];

int read()
{
    int x = 0, k = 1;
    char c; c = getchar();
    while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar(); }
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * k;
}

struct edge
{
    int cnp, to[maxn], last[maxn], head[maxn];
    edge() { cnp = 2; }
    void add(int u, int v)
    { to[cnp] = v, last[cnp] = head[u], head[u] = cnp ++; }
}E1, G;

struct edge1
{
    int u, v;
    edge1(int _u = 0, int _v = 0) { u = _u, v = _v; }
    friend bool operator <(const edge1& a, const edge1& b)
    { return (dfn[a.u] != dfn[b.u]) ? dfn[a.u] < dfn[b.u] : dfn[a.v] < dfn[b.v]; }
}e[maxn];

struct node 
{ 
    int u, num; 
    node(int _u = 0, int _num = 0)
    { u = _u, num = _num; }
};

struct node1
{
    int sum, ls, rs;
}T[maxn * 28];

void dfs(int u)
{
    gra[u][0] = fa[u], dfn[u] = ++ timer; size[u] = 1;
    for(int i = 1; i < CNST; i ++) gra[u][i] = gra[gra[u][i - 1]][i - 1];
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i];
        dep[v] = dep[u] + 1; dfs(v);
        size[u] += size[v];
    }
}

int dfs2(int u)
{
    int lim = 0;
    for(int i = G.head[u]; i; i = G.last[i]) 
    {
        if(u != G.to[i]) lim = (dep[lim] < dep[G.to[i]]) ? lim : G.to[i];
        if(!lim && u != G.to[i]) lim = G.to[i];
    }
    for(int i = E1.head[u]; i; i = E1.last[i])
    {
        int v = E1.to[i];
        int t = dfs2(v); 
        if(u != t && t) lim = (dep[lim] < dep[t]) ? lim : t;
        if(!lim && u != t) lim = t;
    }
    Mul[u][0] = lim;
    return lim;
}

void dfs3(int u)
{
    for(int i = 1; i < CNST; i ++)
        Mul[u][i] = Mul[Mul[u][i - 1]][i - 1];
    for(int i = E1.head[u]; i; i = E1.last[i])
        dfs3(E1.to[i]);
}

int LCA(int u, int v)
{
    if(dep[u] < dep[v]) swap(u, v);
    for(int i = CNST - 1; ~i; i --) 
        if(dep[gra[u][i]] >= dep[v]) u = gra[u][i];
    for(int i = CNST - 1; ~i; i --)
        if(gra[u][i] != gra[v][i]) u = gra[u][i], v = gra[v][i];
    return (u == v) ? u : gra[u][0];
}

node Jump(int u, int fu)
{
    bool flag = 0; int cnt = 0; 
    if(dep[u] == dep[fu]) flag = 1;
    for(int i = CNST - 1; ~i; i --)
    {
        if(!Mul[u][i]) continue;
        if(dep[Mul[u][i]] <= dep[fu]) flag = 1;
        if(dep[Mul[u][i]] > dep[fu]) u = Mul[u][i], cnt += bits[i];
    }
    if(!flag) return node(-1, 0);
    else return node(u, cnt);
}

void Ins(int &now, int pre, int l, int r, int x)
{
    now = ++ tot; T[now] = T[pre]; T[now].sum ++;
    if(l == r) return; 
    int mid = (l + r) >> 1;
    if(x <= mid) Ins(T[now].ls, T[pre].ls, l, mid, x);
    else Ins(T[now].rs, T[pre].rs, mid + 1, r, x);
}

bool Query(int now, int pre, int l, int r, int L, int R)
{
    if(!now) return 0;
    if(L == l && R == r) return (T[now].sum - T[pre].sum);
    if(L > r || R < l) return 0;
    int mid = (l + r) >> 1;
    if(R <= mid) return Query(T[now].ls, T[pre].ls, l, mid, L, R);
    else if(L > mid) return Query(T[now].rs, T[pre].rs, mid + 1, r, L, R);
    else
    {
        if(Query(T[now].ls, T[pre].ls, l, mid, L, mid)) return 1;
        else return Query(T[now].rs, T[pre].rs, mid + 1, r, mid + 1, R);
    }
}

int main()
{
    n = read();
    bits[0] = 1; for(int i = 1; i < CNST; i ++) bits[i] = bits[i - 1] << 1;
    for(int i = 2; i <= n; i ++)
    {
        fa[i] = read();
        E1.add(fa[i], i);
    }
    dep[1] = 1; dfs(1);
    m = read();
    for(int i = 1; i <= m; i ++)
    {
        int u = read(), v = read(), lca = LCA(u, v);
        G.add(u, lca); G.add(v, lca);
        e[i] = edge1(u, v);
    }
    dfs2(1); dfs3(1);
    sort(e + 1, e + 1 + m); int last = 0; root[0] = tot = 1;
    for(int i = 1; i <= m; i ++)
    {
        while(last < dfn[e[i].u] - 1) T[root[last + 1] = ++ tot] = T[root[last]], last ++;
        Ins(root[dfn[e[i].u]], root[last], 1, n, dfn[e[i].v]);
        if(last != dfn[e[i].u]) last ++;
    }
    while(last + 1 <= n) T[root[last + 1] = ++ tot] = T[root[last]], last ++;
    q = read();
    for(int i = 1; i <= q; i ++)
    {
        int u = read(), v = read(), lca = LCA(u, v);
        node a = Jump(u, lca), b = Jump(v, lca); 
        int fu = a.u, fv = b.u;
        if(fu == -1 || fv == -1) { printf("-1
"); continue; }
        if(u == lca || v == lca) { printf("%d
", a.num + b.num + 1); continue; }
        int l1 = dfn[fu], r1 = dfn[fu] + size[fu] - 1;
        int l2 = dfn[fv], r2 = dfn[fv] + size[fv] - 1;
        int x = Query(root[r1], root[l1 - 1], 1, n, l2, r2);
        int y = Query(root[r2], root[l2 - 1], 1, n, l1, r1);
        
        printf("%d
", a.num + b.num + 1 + !(x || y));
    }
    return 0;
}