在树的给定Prufer序列中以质数打印节点

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：

1. 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 [] 。
2. 最初，用1填充度数组。
3. 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。
4. 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。
5. 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