给定一棵树的Prufer序列,任务是在该树中打印带有素数的节点。
例子:
Input: arr[] = {4, 1, 3, 4}
Output: 1 3 4
Explanation:
The tree is:
2----4----3----1----5
|
6
Hence, the degree of 1, 3 and 4
are 2, 2 and 3 respectively
which are prime.
Input: a[] = {1, 2, 2}
Output: 1 2
方法:
- 因为如果N是节点数,则prufer序列的长度为N – 2 。因此,创建一个比Prufer序列的长度大2的数组degree [] 。
- 最初,用1填充度数组。
- 对Prufer序列进行迭代,并增加度表中每个元素的频率。此方法之所以有效,是因为Prufer序列中节点的频率比树中的度数小1。
- 此外,要检查节点度数是否为质数,我们将使用过筛网。创建一个筛子,这将有助于我们确定度数是否在O(1)时间内为素数。
- 如果节点具有素数,则打印该节点号。
下面是上述方法的实现:
C++
// C++ implementation to print the
// nodes with prime degree from the
// given prufer sequence
#include
using namespace std;
// Function to create Sieve
// to check primes
void SieveOfEratosthenes(
bool prime[], int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2; i <= p_size;
i += p)
prime[i] = false;
}
}
}
// Function to print the nodes with
// prime degree in the tree
// whose Prufer sequence is given
void PrimeDegreeNodes(int prufer[], int n)
{
int nodes = n + 2;
bool prime[nodes + 1];
memset(prime, true, sizeof(prime));
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
int degree[n + 2 + 1];
// Initially let all the degrees be 1
for (int i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for (int i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for (int i = 1; i <= nodes; i++) {
if (prime[degree[i]]) {
cout << i << " ";
}
}
}
// Driver Code
int main()
{
int a[] = { 4, 1, 3, 4 };
int n = sizeof(a) / sizeof(a[0]);
PrimeDegreeNodes(a, n);
return 0;
}
Java
// Java implementation to print the
// nodes with prime degree from the
// given prufer sequence
import java.util.*;
class GFG{
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(
boolean prime[], int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for (int p = 2; p * p <= p_size; p++) {
// If prime[p] is not changed,
// then it is a prime
if (prime[p]) {
// Update all multiples of p,
// set them to non-prime
for (int i = p * 2; i <= p_size;
i += p)
prime[i] = false;
}
}
}
// Function to print the nodes with
// prime degree in the tree
// whose Prufer sequence is given
static void PrimeDegreeNodes(int prufer[], int n)
{
int nodes = n + 2;
boolean []prime = new boolean[nodes + 1];
Arrays.fill(prime, true);
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
int []degree = new int[n + 2 + 1];
// Initially let all the degrees be 1
for (int i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for (int i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for (int i = 1; i <= nodes; i++) {
if (prime[degree[i]]) {
System.out.print(i+ " ");
}
}
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 4, 1, 3, 4 };
int n = a.length;
PrimeDegreeNodes(a, n);
}
}
// This code contributed by Princi Singh
Python3
# Python3 implementation to print the
# nodes with prime degree from the
# given prufer sequence
# Function to create Sieve
# to check primes
def SieveOfEratosthenes(prime, p_size):
# False here indicates
# that it is not prime
prime[0] = False
prime[1] = False
p = 2
while (p * p <= p_size):
# If prime[p] is not changed,
# then it is a prime
if (prime[p]):
# Update all multiples of p,
# set them to non-prime
for i in range(p * 2, p_size + 1, p):
prime[i] = False
p += 1
# Function to print the nodes with
# prime degree in the tree
# whose Prufer sequence is given
def PrimeDegreeNodes(prufer, n):
nodes = n + 2
prime = [True] * (nodes + 1)
SieveOfEratosthenes(prime, nodes + 1)
# Hash-table to mark the
# degree of every node
degree = [0] * (n + 2 + 1);
# Initially let all the degrees be 1
for i in range(1, nodes + 1):
degree[i] = 1;
# Increase the count of the degree
for i in range(0, n):
degree[prufer[i]] += 1
# Print the nodes with prime degree
for i in range(1, nodes + 1):
if prime[degree[i]]:
print(i, end = ' ')
# Driver Code
if __name__=='__main__':
a = [ 4, 1, 3, 4 ]
n = len(a)
PrimeDegreeNodes(a, n)
# This code is contributed by rutvik_56
C#
// C# implementation to print the
// nodes with prime degree from the
// given prufer sequence
using System;
class GFG{
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(bool []prime,
int p_size)
{
// False here indicates
// that it is not prime
prime[0] = false;
prime[1] = false;
for(int p = 2; p * p <= p_size; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p])
{
// Update all multiples of p,
// set them to non-prime
for(int i = p * 2; i <= p_size;
i += p)
prime[i] = false;
}
}
}
// Function to print the nodes with
// prime degree in the tree
// whose Prufer sequence is given
static void PrimeDegreeNodes(int []prufer, int n)
{
int nodes = n + 2;
bool []prime = new bool[nodes + 1];
for(int i = 0; i < prime.Length; i++)
prime[i] = true;
SieveOfEratosthenes(prime, nodes + 1);
// Hash-table to mark the
// degree of every node
int []degree = new int[n + 2 + 1];
// Initially let all the degrees be 1
for(int i = 1; i <= nodes; i++)
degree[i] = 1;
// Increase the count of the degree
for(int i = 0; i < n; i++)
degree[prufer[i]]++;
// Print the nodes with prime degree
for(int i = 1; i <= nodes; i++)
{
if (prime[degree[i]])
{
Console.Write(i + " ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 4, 1, 3, 4 };
int n = a.Length;
PrimeDegreeNodes(a, n);
}
}
// This code is contributed by 29AjayKumar
输出:
1 3 4
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