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📜  计算给定树中权重为完美数的节点

📅  最后修改于: 2021-04-22 08:02:53             🧑  作者: Mango

给定一棵树,以及所有节点的权重,任务是计算权重为Perfect number的节点的数量

例子:

方法:
为了解决此问题,我们在树上执行深度优先搜索(DFS)遍历,并针对每个节点检查其权重是否为“完美数”。每当获得这样的权重时,我们就不断增加计数器。整个树遍历完成后该计数器的最终值就是答案。
下面是上述方法的实现:

C++
// C++ implementation to Count the nodes in the
// given tree whose weight is a Perfect Number
 
#include 
using namespace std;
 
int ans = 0;
vector graph[100];
vector weight(100);
 
// Function that returns true if n is perfect
bool isPerfect(long long int n)
{
    // Variable to store sum of divisors
    long long int sum = 1;
 
    // Find all divisors and add them
    for (long long int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
 
    // Check if sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Function to perform dfs
void dfs(int node, int parent)
{
 
    // If weight of the current node
    // is a perfect number
    if (isPerfect(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 to Count the nodes in the
// given tree whose weight is a Perfect Number
 
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 perfect
static boolean isPerfect(int n)
{
    // Variable to store sum of divisors
    int sum = 1;
 
    // Find all divisors and add them
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
 
    // Check if sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
 
    return false;
}
 
// Function to perform dfs
static void dfs(int node, int parent)
{
 
    // If weight of the current node
    // is a perfect number
    if (isPerfect(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 < graph.length; 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 contributed by Princi Singh


Python3
# Python3 implementation to
# Count the Nodes in the given
# tree whose weight is a Perfect
# Number
 
graph = [[] for i in range(100)]
weight = [0] * 100
ans = 0
 
# Function that returns
# True if n is perfect
def isPerfect(n):
   
    # Variable to store
    # sum of divisors
    sum = 1;
 
    # Find all divisors
    # and add them
    i = 2;
     
    while(i * i < n):
        if (n % i == 0):
            if (i * i != n):
                sum = sum + i + n / i;
            else:
                sum = sum + i;
        i += 1;
 
    # Check if sum of divisors
    # is equal to n, then n is
    # a perfect number
    if (sum == n and n != 1):
        return True;
 
    return False;
 
# Function to perform dfs
def dfs(Node, parent):
   
    # If weight of the current
    # Node is a perfect number
    global ans;
     
    if (isPerfect(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 29AjayKumar


C#
// C# implementation to count the
// nodes in the given tree whose
// weight is a Perfect Number
using System;
using System.Collections.Generic;
 
class GFG{
 
static int ans = 0;
static List []graph = new List[100];
static int []weight = new int[100];
 
// Function that returns true
// if n is perfect
static bool isPerfect(int n)
{
     
    // Variable to store sum of
    // divisors
    int sum = 1;
 
    // Find all divisors and add them
    for(int i = 2; i * i <= n; i++)
    {
       if (n % i == 0)
       {
           if (i * i != n)
               sum = sum + i + n / i;
           else
               sum = sum + i;
       }
    }
 
    // Check if sum of divisors is equal
    // to n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
    return false;
}
 
// Function to perform dfs
static void dfs(int node, int parent)
{
 
    // If weight of the current node
    // is a perfect number
    if (isPerfect(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 < graph.Length; i++)
       graph[i] = new List();
         
    // 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 amal kumar choubey


输出:
1


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,树的每个节点都处理一次,因此,如果树中总共有N个节点,则由于dfs而导致的复杂度为O(N)。另外,在处理每个节点时,为了检查节点值是否为完美数,将调用isPerfect(V)函数(其中V为节点的权重),该函数的复杂度为O(logV) ,因此对于每个节点,都会增加O(logV)的复杂度。因此,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。