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];
int t, dad[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;
dad[y] = x;
graph[x].push_back(y);
}
for(int i = 1; i <= n; ++i)
if(!dad[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 {
BufferedReader in;
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();
in = new BufferedReader(new InputStreamReader(System.in));
out = new PrintWriter(System.out);
// in = new BufferedReader(new FileReader("src/input.txt"));
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);
}
}
String readString() throws IOException {
while (!tok.hasMoreTokens()) {
tok = new StringTokenizer(in.readLine());
}
return tok.nextToken();
}
int readInt() throws IOException {
return Integer.parseInt(readString());
}
long readLong() throws IOException {
return Long.parseLong(readString());
}
double readDouble() throws IOException {
return Double.parseDouble(readString());
}
Edge[] first;
FenwickTree sum;
long result;
void solve() throws IOException {
int n = readInt();
int k = readInt();
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)
def add(x, v):
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)
add(x, 1)
vis[x] = 1
if nx[x]:
st.append(nx[x][-1])
nx[x].pop()
else:
st.pop()
add(x, -1)
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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