计算权重为完美平方的节点

给定一棵树，以及所有节点的权重，任务是计算权重为完美Square的节点的数量。
例子：

Input: 
 

Output: 3 
Only the weights of nodes 1, 4 and 5 are perfect squares. 
 

方法：在树上执行dfs，对于每个节点，检查其权重是否为理想平方。
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 a perfect square
bool isPerfectSquare(int n)
{
    double x = sqrt(n);
    if (floor(x) != ceil(x))
        return false;
    return true;
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(weight[node]))
        ans += 1;
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    int x = 15;
 
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
 
    // Edges of the tree
    graph[1].push_back(2);
    graph[2].push_back(3);
    graph[2].push_back(4);
    graph[1].push_back(5);
    graph[5].push_back(6);
 
    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 a perfect square
static boolean isPerfectSquare(int n)
{
    double x = Math.sqrt(n);
    if (Math.floor(x) != Math.ceil(x))
        return false;
    return true;
}
  
// Function to perform dfs
static void dfs(int node, int parent)
{
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(weight[node]))
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
public static void main(String[] args)
{
    int x = 15;
    for (int i = 0; i < 100; i++)
        graph[i] = new Vector<>();
     
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
  
    // Edges of the tree
    graph[1].add(2);
    graph[2].add(3);
    graph[2].add(4);
    graph[1].add(5);
    graph[5].add(6);
  
    dfs(1, 1);
  
    System.out.print(ans);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
from math import *
ans = 0
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function that returns true
# if n is a perfect square
def isPerfectSquare(n):
    x = sqrt(n)
    if (floor(x) != ceil(x)):
        return False
    return True
 
# Function to perform dfs
def dfs(node, parent):
    global ans
     
    # If weight of the current node
    # is a perfect square
    if (isPerfectSquare(weight[node])):
        ans += 1;
     
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
 
# Driver code
 
x = 15
 
# Weights of the node
weight[1] = 4
weight[2] = 5
weight[3] = 3
weight[4] = 25
weight[5] = 16
weight[6] = 30
 
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
 
dfs(1, 1)
print(ans)
 
# This code is contributed by SHUBHAMSINGH10

C#

// C# program for the above 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 a perfect square
static bool isPerfectSquare(int n)
{
    double x = Math.Sqrt(n);
     
    if (Math.Floor(x) != Math.Ceiling(x))
        return false;
         
    return true;
}
 
// Function to perform dfs
static void dfs(int node, int parent)
{
     
    // If weight of the current node
    // is a perfect square
    if (isPerfectSquare(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)
{
    //int x = 15;
    for(int i = 0; i < 100; i++)
        graph[i] = new ArrayList();
     
    // Weights of the node
    weight[1] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
 
    // Edges of the tree
    graph[1].Add(2);
    graph[2].Add(3);
    graph[2].Add(4);
    graph[1].Add(5);
    graph[5].Add(6);
 
    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 a perfect square
    function isPerfectSquare(n)
    {
        let x = Math.sqrt(n);
        if (Math.floor(x) != Math.ceil(x))
            return false;
        return true;
    }
     
    // Function to perform dfs
    function dfs(node,parent)
    {
        // If weight of the current node
        // is a perfect square
        if (isPerfectSquare(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] = 4;
    weight[2] = 5;
    weight[3] = 3;
    weight[4] = 25;
    weight[5] = 16;
    weight[6] = 30;
       
   
    // Edges of the tree
    graph[1].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);
    graph[5].push(6);
     
    dfs(1, 1);
   
    document.write( ans);
     
    // This code is contributed by unknown2108
     

输出：

3

复杂度分析：

• 时间复杂度： O(N * logV)其中V是树中节点的最大权重。
  在DFS中，树的每个节点都被处理一次，因此对于树中的N个节点，由于DFS而导致的复杂度为O(N)。同样，在处理每个节点时，为了检查节点值是否为完美平方，将调用内置的sqrt(V)，其中V是节点的权重，此函数的复杂度为O(log V)。因此，对于每个节点，都会增加O(log V)的复杂度。因此，总时间复杂度为O(N * logV)。
• 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。