计算给定树中权重为素数的节点

给定一棵树，以及所有节点的权重，任务是计算权重为素数的节点数。
例子：

Input: 
 

Output: 2 
Only the weights of the nodes 1 and 3 are prime. 
 

方法：在树上执行 dfs，对于每个节点，检查它的权重是否为素数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int ans = 0;
 
vector graph[100];
vector weight(100);
 
// Function that returns true
// if n is prime
bool isprime(int n)
{
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of node is prime or not
    if (isprime(weight[node]))
        ans += 1;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
 
    dfs(1, 1);
 
    cout << ans;
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG{
  
static int ans = 0;
 
static Vector[] graph = new Vector[100];
static int[] weight = new int[100];
  
// Function that returns true
// if n is prime
static boolean isprime(int n)
{
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
    return true;
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
    // If weight of node is prime or not
    if (isprime(weight[node]))
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
public static void main(String[] args)
{
    for (int i = 0; i < 100; i++)
        graph[i] = new Vector<>();
     
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
  
    // Edges of the tree
    graph[1].add(2);
    graph[2].add(3);
    graph[2].add(4);
    graph[1].add(5);
  
    dfs(1, 1);
  
    System.out.print(ans);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
ans = 0
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function that returns true
# if n is prime
def isprime(n):
    i = 2
    while(i * i <= n):
        if (n % i == 0):
            return False
        i += 1
    return True
 
# Function to perform dfs
def dfs(node, parent):
    global ans
     
    # If weight of the current node is even
    if (isprime(weight[node])):
        ans += 1;
     
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
# Driver code
 
# Weights of the node
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
 
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
 
dfs(1, 1)
print(ans)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;
 
class GFG{
     
static int ans = 0;
static ArrayList[] graph = new ArrayList[100];
static int[] weight = new int[100];
 
// Function that returns true
// if n is prime
static bool isprime(int n)
{
    for(int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return false;
             
    return true;
}
 
// Function to perform dfs
static void dfs(int node, int parent)
{
     
    // If weight of node is prime or not
    if (isprime(weight[node]))
        ans += 1;
 
    foreach(int to in graph[node])
    {
        if (to == parent)
            continue;
             
        dfs(to, node);
    }
}
     
// Driver Code
public static void Main(string[] args)
{
    for(int i = 0; i < 100; i++)
        graph[i] = new ArrayList();
     
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
 
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
 
    dfs(1, 1);
 
    Console.Write(ans);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript implementation of the approach
     
    let ans=0;
     
    let graph = new Array(100);
     
     
    let weight = new Array(100);
    for(let i=0;i<100;i++)
    {
        graph[i]=[];
        weight[i]=0;
    }
     
    // Function that returns true
    // if n is prime
    function isprime(n)
    {
        for (let i = 2; i * i <= n; i++)
            if (n % i == 0)
                return false;
        return true;
    }
     
    // Function to perform dfs
    function dfs(node,parent)
    {
         // If weight of node is prime or not
        if (isprime(weight[node]))
            ans += 1;
        for(let to=0;to<graph[node].length;to++)
        {
            if(graph[node][to] == parent)
                continue
            dfs(graph[node][to], node);  
        }
         
    }
     
    // Driver code
     
    x = 15;
   
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
       
   
    // Edges of the tree
    graph[1].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
     
     
    dfs(1, 1);
   
    document.write( ans);
     
    // This code is contributed by unknown2108
     

输出：

2

复杂度分析：

• 时间复杂度： O(N*sqrt(V))，其中 V 是给定树中节点的最大权重。
  在 DFS 中，树的每个节点都被处理一次，因此当树中总共有 N 个节点时，由于 DFS 的复杂性是 O(N)。此外，在处理每个节点时，为了检查节点值是否为素数，正在运行一个直到 sqrt(V) 的循环，其中 V 是节点的权重。因此，对于每个节点，都会增加 O(sqrt(V)) 的复杂度。因此，时间复杂度为 O(N*sqrt(V))。
• 辅助空间： O(1)。
  不需要任何额外的空间，因此空间复杂度是恒定的。

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