CSU 1414:query on a tree

Preprocessing the number of sub nodes of each node sons, then the query to X can be obtained by sons[x]-Sigma (Sons[v) (V is the point to X distance to D)

How to find these v quickly? Notice that the point from X to D is definitely on the same level as the tree ....

You can record each node timestamp while the tree is DFS and save the nodes of each layer. Then maintain a prefix for each layer and if V is x under the face node then V's timestamp must be within the range of X, so that the prefix and range can be determined by two points ....

1414:query on a tree
Time Limit:3 Sec Memory limit:128 MB
submit:78 solved:23
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Description
You are are given a rooted tree with N nodes indexed by 1, 2, ..., N, and the root is indexed by 1. We'll 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 this distance between it and X is less than D.
Assume the length of each edge is 1.
Input
The ' the ' I line contains ' an integer t (T > 0), giving the number of test cases.
For the all test case, the ' the ' the ' 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 O f 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 on subtree rooted at x, such this distance between it and X is Les S 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

The eighth session of university Students ' program design competition
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#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include &l  
      
T;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<s  
            z;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:" <&  
    lt;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;  
    The left and right interval 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", &AMP;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; }