# Similar Pair

### Problem Statement :

```A pair of nodes, , is a similar pair if the following conditions are true:

node  is the ancestor of node
Given a tree where each node is labeled from  to , find the number of similar pairs in the tree.

Function Description

Complete the similarPair function in the editor below. It should return an integer that represents the number of pairs meeting the criteria.

similarPair has the following parameter(s):

n: an integer that represents the number of nodes
k: an integer
edges: a two dimensional array where each element consists of two integers that represent connected node numbers
Input Format

The first line contains two space-separated integers  and , the number of nodes and the similarity threshold.
Each of the next  lines contains two space-separated integers defining an edge connecting nodes  and , where node  is the parent to node .```

### Solution :

```                            ```Solution in C :

In  C :

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#include "math.h"

typedef struct Node
{
struct Node *parent;
struct Node *peer_next;
struct Node *child_list;
int     val;
struct Node *hash_next;
}Node;

unsigned long long int count;
unsigned int n,T,size;
Node **hash;
Node *root=NULL;

unsigned int diff(int a, int b)
{
if(a>b) return (a-b);
else    return (b-a);
}

void countup(Node *x)
{
int i,val;
if(!x || !x->parent) return;
if((n-T) < size)
{
count+=size;
for(i=0;i<(((x->val-1)>T)?(x->val-1-T):0); i++)
if(hash[i]) count--;
for(i=(((x->val+T)>n)?n:(x->val+T));i<n; i++)
if(hash[i]) count--;
}
else if(T > size)
{
val=x->val;
x=x->parent;
while(x)
{
if(diff(val,x->val) <= T) count++;
x=x->parent;
}
}
else
{
for(i=((x->val-1)>T)?(x->val-1-T):0; i<(((x->val+T)>n)?n:(x->val+T)); i++)
{
if(hash[i])
{
//printf("%2d, 0x%x\n",i,hash[i]);
count++;
}
}
}
}

void solve()
{
Node *tmp=root;
Node *tmp1;
int i;
for(i=0;i<n;i++) hash[i]=NULL;
size=0;
while(tmp)
{
while(tmp->child_list)
{
hash[(tmp->val-1)%n]=tmp;
size++;
tmp=tmp->child_list;
}

countup(tmp);
tmp1=tmp;
tmp=tmp->parent;
if(tmp)// && (tmp->child_list == tmp1))
{
hash[(tmp->val-1)%n]=NULL;
size--;
tmp->child_list=tmp1->peer_next;
}
//printf("node = %3d (count = %d)\n",tmp1->val,count);
free(tmp1);
}
}

Node* allocate(unsigned int val)
{
Node *node=malloc(sizeof(Node));
memset(node,0,sizeof(Node));
node->val=val;
return node;
}
Node* insert(unsigned int val)
{
Node *tmp=hash[val%n];
if(!tmp)
{
return (hash[val%n]=allocate(val));
}
while(tmp)
{
if(tmp->val==val) return tmp;
if(!tmp->hash_next)
break;
tmp=tmp->hash_next;
}
return (tmp->hash_next=allocate(val));
}

void connect(Node *parent, Node *child)
{
if(!parent || !child) return;
/*if(!parent->child_list)
parent->child_list=child;
else
{
Node *peer=parent->child_list;
while(peer->peer_next) peer=peer->peer_next;
peer->peer_next=child;
}*/
child->peer_next=parent->child_list;
parent->child_list=child;

child->parent=parent;
}

void build(){

int i,a,b;
Node *parent,*child;
for(i=0;i<n-1;i++)
{
scanf("%d %d",&a,&b);
parent=insert(a);
child=insert(b);
//printf("%d %d\n",parent->val,child->val);
connect(parent,child);
/*if(!parent->parent)
root=parent;*/
}
root=hash[1];
while(root && root->parent) root=root->parent;
}

void print(Node *node, int level)
{
int i=level;
if(!node) return;
while(i--) printf("  ");
printf("%d (%d)\n",node->val,node->parent?node->parent->val:0);
node=node->child_list;
while(node)
{
print(node,level+1);
node=node->peer_next;
}
}

int main(){
count=0;
scanf("%d %d",&n,&T);
hash=malloc(n*sizeof(Node*));
memset(hash,0,n*sizeof(Node*));
if (!hash) return -1;
build();
//print(root, 0);
solve();
printf("%llu\n",count);
return 0;
}```
```

```                        ```Solution in C++ :

In   C++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int n, aib[200005];

inline int lsb(int & x){
return x & -x;
}

void update(int val, int pos){
for(int i = pos; i <= n * 2; i += lsb(i))
aib[i] += val;
}

int query(int pos){
int rval = 0;
for(int i = pos; i > 0; i -= lsb(i))
rval += aib[i];
return rval;
}

vector<int> graph[100005];
long long ans;

void dfs(int x){
ans += (long long)query(x + t) - query(x - t - 1);
update(1, x);
for(int i = 0; i < graph[x].size(); ++i)
dfs(graph[x][i]);
update(-1, x);
}

int main() {
cin >> n >> t;
for(int i = 1; i < n; ++i){
int x, y;
cin >> x >> y;
graph[x].push_back(y);
}
for(int i = 1; i <= n; ++i)
dfs(i);
cout << ans;
return 0;
}```
```

```                        ```Solution in Java :

In  Java  :

import java.awt.Point;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import static java.lang.Math.*;

public class Solution implements Runnable {

PrintWriter out;
StringTokenizer tok = new StringTokenizer("");

public static void main(String[] args) {
new Thread(null, new Solution(), "", 256 * (1L << 20)).start();
}

public void run() {
try {
long t1 = System.currentTimeMillis();
out = new PrintWriter(System.out);

Locale.setDefault(Locale.US);
solve();
in.close();
out.close();
long t2 = System.currentTimeMillis();
System.err.println("Time = " + (t2 - t1));
} catch (Throwable t) {
t.printStackTrace(System.err);
System.exit(-1);
}
}

while (!tok.hasMoreTokens()) {
}
}

}

}

}
Edge[] first;
FenwickTree sum;
long result;

void solve() throws IOException {
first = new Edge[n];
boolean[] root = new boolean[n];
Arrays.fill(root, true);
for (int i = 0; i < n - 1; i++) {
int from = readInt() - 1;
int to = readInt() - 1;
root[to] = false;
first[from] = new Edge(from, to, first[from]);
}
sum = new FenwickTree(n);
result = 0;
for (int i = 0; i < n; i++) {
if (root[i]) {
dfs(i, k);
break;
}
}
out.println(result);
}

void dfs(int x, int k)
{
result += sum.find(x + k) - sum.find(x - k - 1);
sum.increase(x, +1);
for (Edge edge = first[x]; edge != null; edge = edge.next)
{
dfs(edge.b, k);
}
sum.increase(x, -1);
}

class Edge {

int a;
int b;
Edge next;

Edge(int a, int b, Edge next) {
this.a = a;
this.b = b;
this.next = next;
}
}

class FenwickTree {

private int[] sum;

FenwickTree(int size) {
sum = new int[size + 10];
}

private int prev(int x) {
return x & (x - 1);
}

private int next(int x) {
return 2 * x - prev(x);
}

void increase(int id, int value) {
id++;
while (id < sum.length) {
sum[id] += value;
id = next(id);
}
}

long find(int id) {
id++;
id = Math.min(sum.length - 1, id);
long res = 0;
if (id <= 0) {
return 0;
}
while (id > 0) {
res += sum[id];
id = prev(id);
}
return res;
}
}
}```
```

```                        ```Solution in Python :

In  Python3 :

import resource
import sys
sys.setrecursionlimit(2000000)

x += 1
while x <= n:
a[x] += v
x += x & -x

def que(x):
x += 1
if x <= 0:
return 0
ret = 0
x = min(n, x)
while x > 0:
ret += a[x]
x -= x & -x
return ret

st = []
vis = {}
def dfs(x):

global ans
st.append(x)
while st:
x = st[-1]
if not x in vis:
ans += que(x + T) - que(x - T - 1)
vis[x] = 1
if nx[x]:
st.append(nx[x][-1])
nx[x].pop()
else:
st.pop()

n, T = (int(x) for x in input().split())
a = [0 for i in range(4 * n)]
nx = [[] for i in range(n)]
pre = [-1 for i in range(n)]
for i in range(n - 1):
s, e = (int(x) - 1 for x in input().split())
nx[s].append(e)
pre[e] = s

s = 1
while pre[s] != -1:
s = pre[s]
ans = 0
dfs(s)
print(ans)```
```

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