CSU 1414: Query on a Tree

预处理每个结点的子结点的个数sons , 则对x的询问可由sons[x]- sigma( sons[v] ) (v是到x距离为d的点)得到

怎么快速的找到这些v呢? 注意到距离x为d的点肯定在树的同一层....

可以对树进行dfs时记录每个结点时间戳的同时把每一层的结点保存下来,然后对每一层维护一个前缀和 如果v是x下面子结点那么v的时间戳肯定在x的范围内,这样就可以二分確定出前缀和的范围了.....

1414: Query on a Tree
Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 78  Solved: 23
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Description
You are given a rooted tree with N nodes indexed by 1, 2, ..., N, and the root is indexed by 1. We will ask you to perform some queries of the following form:
x d : Ask for how many nodes there are in the subtree rooted at x, such that the distance between it and x is less than d.
Assume the length of each edge is 1.
Input
The first line contains an integer T (T > 0), giving the number of test cases.
For each test case, the first line contains an integer N (2 ≤ N ≤ 105). The next line contains N - 1 integers f2, f3, ..., fN (1 ≤ f2, f3, ..., fn ≤ N), where fi (2 ≤ i ≤ N) denotes the father of i is fi. Then follows a line with an integer M (1 ≤ M ≤ 105) giving the number of queries. Then follow M lines with two integers x, d (1 ≤ x, d ≤ N), giving the M queries.
Output
For each query, output how many nodes there are in the subtree rooted at x, such that the distance between it and x is less than d.
Sample Input

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1
9
1 2 2 4 4 2 1 8
6
1 1
1 2
1 3
2 3
2 4
3 2

Sample Output

1
3
7
6
6
1

HINT

Source

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>  

using namespace std;  

const int maxn=110000;  

struct Edge
{
    int to,next;
}edge[maxn];
int Adj[maxn],Size;
vector<int> eg[maxn];
vector<int> pre_sum[maxn];
int dfn[maxn][2],sons[maxn],max_deep[maxn],Deep[maxn],id[maxn],ti,maxdeep,n;  

void init(int n)
{
    Size=0;
    memset(Adj,-1,sizeof(Adj));
    memset(dfn,0,sizeof(dfn));
    memset(sons,0,sizeof(sons));
    memset(Deep,0,sizeof(Deep));
    memset(id,0,sizeof(id));
    memset(max_deep,0,sizeof(max_deep));
    ti=0;maxdeep=0;
    for(int i=0;i<=n+20;i++)
    {
        pre_sum[i].clear(),pre_sum[i].push_back(0);
        eg[i].clear(),eg[i].push_back(0);
    }
}  

void Add_Edge(int u,int v)
{
    edge[Size].to=v;
    edge[Size].next=Adj[u];
    Adj[u]=Size++;
}  

void DFS(int u,int deep)
{
    dfn[u][0]=++ti;
    maxdeep=max(maxdeep,deep);
    eg[deep].push_back(u);
    int maxd=0;
    for(int i=Adj[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        DFS(v,deep+1);
        maxd=max(maxd,max_deep[v]);
    }
    max_deep[u]=maxd+1;
    Deep[u]=deep;
    dfn[u][1]=ti;
    sons[u]=dfn[u][1]-dfn[u][0]+1;
}  

void prefix_sum()
{
    for(int i=1;i<=maxdeep;i++)
    {
        for(int j=1,sz=eg[i].size();j<sz;j++)
        {
            id[eg[i][j]]=j;
            pre_sum[i].push_back( pre_sum[i][j-1]+sons[eg[i][j]] );
        }
    }
}  

void Debug()
{
    for(int i=1;i<=n;i++)
    {
        cout<<i<<" st: "<<dfn[i][0]<<" et: "<<dfn[i][1]<<"   sons: "<<sons[i]<<endl;
    }
    cout<<"maxdeep: "<<maxdeep<<endl;
    for(int i=1;i<=maxdeep;i++)
    {
        cout<<i<<": "<<endl;
        for(int j=0,sz=eg[i].size();j<sz;j++)
        {
            cout<<eg[i][j]<<",";
        }
        cout<<endl;
    }
    cout<<"id: \n";
    for(int i=1;i<=maxdeep;i++)
    {
        cout<<i<<": "<<endl;
        for(int j=0,sz=eg[i].size();j<sz;j++)
        {
            cout<<id[eg[i][j]]<<",";
        }
        cout<<endl;
    }
    cout<<"prefix_sum: \n";
    for(int i=1;i<=maxdeep;i++)
    {
        cout<<i<<": "<<endl;
        for(int j=0,sz=eg[i].size();j<sz;j++)
        {
            cout<<pre_sum[i][j]<<",";
        }
        cout<<endl;
    }
    cout<<"max_deep: ";
    for(int i=1;i<=n;i++)
    {
        cout<<i<<" : "<<max_deep[i]<<endl;
    }
}  

void solve(int x,int d)
{
    int nowd=Deep[x];
    if(d>=max_deep[x])
    {
        printf("%d\n",sons[x]);
        return ;
    }
    int qd=nowd+d;
    ///二分左右区间
    int L=0,R=0,sz=eg[qd].size();
    int low,mid,high;
    ///....get left pos
    low=0,high=sz-1;
    while(low<=high)
    {
        mid=(low+high)/2;
        int ps=eg[qd][mid];
        if(dfn[ps][0]>=dfn[x][0])
        {
            L=eg[qd][mid-1]; high=mid-1;
        }
        else low=mid+1;
    }
    ///...get right pos
    low=0,high=sz-1;
    while(low<=high)
    {
        mid=(low+high)/2;
        int ps=eg[qd][mid];
        if(dfn[ps][1]<=dfn[x][1])
        {
            R=ps; low=mid+1;
        }
        else high=mid-1;
    }
    printf("%d\n",sons[x]-pre_sum[qd][id[R]]+pre_sum[qd][id[L]]);
}  

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d",&n);
        init(n+1);
        for(int i=2;i<=n;i++)
        {
            int fa;
            scanf("%d",&fa);
            Add_Edge(fa,i);
        }
        DFS(1,1);
        prefix_sum();
       // Debug();
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int x,d;
            scanf("%d%d",&x,&d);
            solve(x,d);
        }
    }
    return 0;
}

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