Blood Cousins 【CodeForces - 208E】【Dsu on Tree】

题目链接

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  在一个森林中，如果两个节点a和b向上的第p个祖先相同，就称他们为p代表亲。(跟日常生活中有所不同，p不一定是他们的最近公共祖先)。 
  给出一些询问，问v的p代表亲的数量。

  我们肯定是要想办法把节点推到子树的根结点上面去，这样问题一转，就变成了求子结点的问题了，我们可以用倍增的方法，找到对应的祖先节点，然后就是我们将询问放到该节点上面来查询这棵子树即可。

时间复杂度

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, Q, head[maxN], cnt;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
int siz[maxN], root[maxN][20], dfn[maxN], tot, rid[maxN], Wson[maxN], deep[maxN];
void pre_dfs(int u, int fa)
{
    siz[u] = 1; dfn[u] = ++tot; rid[tot] = u;
    int maxx = 0;
    root[u][0] = fa;
    for(int i=0; (1<<(i + 1)) < N; i++) root[u][i + 1] = root[root[u][i]][i];
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        deep[v] = deep[u] + 1;
        pre_dfs(v, u);
        siz[u] += siz[v];
        if(siz[v] > maxx)
        {
            maxx = siz[v];
            Wson[u] = v;
        }
    }
}
inline int fid_father(int u, int lenth)
{
    for(int i=log2(lenth); i>=0; i--)
    {
        if(lenth & (1 << i))
        {
            u = root[u][i];
        }
    }
    return u;
}
struct Question
{
    int id, depth;
    Question(int a=0, int b=0):id(a), depth(b) {}
};
vector<Question> vt[maxN];
int ans[maxN], siz_deep[maxN] = {0};
void dfs(int u, bool keep)
{
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == Wson[u]) continue;
        dfs(v, false);
    }
    if(Wson[u])
    {
        dfs(Wson[u], true);
    }
    for(int i=head[u], v, id; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == Wson[u]) continue;
        for(int j=dfn[v]; j<dfn[v] + siz[v]; j++)
        {
            id = rid[j];
            siz_deep[deep[id]]++;
        }
    }
    int len = (int)vt[u].size();
    Question now;
    for(int i=0; i<len; i++)
    {
        now = vt[u][i];
        ans[now.id] = siz_deep[now.depth] - 1;
    }
    if(!keep)
    {
        for(int i=dfn[u] + 1, v; i<dfn[u] + siz[u]; i++)
        {
            v = rid[i];
            siz_deep[deep[v]] = 0;
        }
    }
    else
    {
        siz_deep[deep[u]]++;
    }
}
inline void init()
{
    cnt = 0;
    for(int i=0; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d", &N);
    init();
    for(int i=1, ff; i<=N; i++)
    {
        scanf("%d", &ff);
        addEddge(ff, i);
    }
    pre_dfs(0, 0);
    scanf("%d", &Q);
    for(int i=1, v, p, u; i<=Q; i++)
    {
        scanf("%d%d", &v, &p);
        u = fid_father(v, p);
        if(u) vt[u].push_back(Question(i, deep[v]));
        else ans[i] = 0;
    }
    dfs(0, false);
    for(int i=1; i<=Q; i++) printf("%d%c", ans[i], i == Q ? '\n' : ' ');
    return 0;
}