Dado un árbol n-ario que tiene Nodes con un peso particular, nuestra tarea es encontrar la suma máxima de la diferencia mínima de divisores de cada Node desde la raíz hasta la hoja.
Ejemplos:
Aporte:
18 / \ 7 15 / \ \ 4 12 2 / 9Salida: 10
Explicación:
La suma máxima está a lo largo del camino 18 -> 7 -> 12 -> 9
Diferencia mínima entre divisores de 18 = 6 – 3 = 3
Diferencia mínima entre divisores de 7 = 7 – 1 = 6
Diferencia mínima entre divisores de 12 = 4 – 3 = 1
Diferencia mínima entre divisores de 9 = 3 – 3 = 0Aporte:
20 / \ 13 14 / \ \ 10 8 26 / 25Salida: 17
Explicación:
La suma máxima está a lo largo del camino 20 -> 14 -> 26
Enfoque:
para resolver el problema mencionado anteriormente, podemos almacenar la diferencia mínima entre los divisores en cada Node en una array utilizando el primer recorrido en profundidad . Ahora, la tarea es encontrar la diferencia mínima entre los divisores.
- Podemos observar que para cualquier número N, el divisor x más pequeño en el rango [√N, N] tendrá la menor diferencia entre x y N/x .
- En cada Node, calcule la diferencia mínima entre los divisores y, finalmente, comience a llenar la array con los resultados de los niños y calcule la suma máxima posible.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
#include <bits/stdc++.h>
using namespace std;
// Array to store the
// result at each node
int sub[100005];
// Function to get minimum difference
// between the divisors of a number
int minDivisorDifference(int n)
{
int num1;
int num2;
// Iterate from square
// root of N to N
for (int i = sqrt(n); i <= n; i++) {
if (n % i == 0) {
num1 = i;
num2 = n / i;
break;
}
}
// return absolute difference
return abs(num1 - num2);
}
// DFS function to calculate the maximum sum
int dfs(vector<int> g[], int u, int par)
{
// Store the min difference
sub[u] = minDivisorDifference(u);
int mx = 0;
for (auto c : g[u]) {
if (c != par) {
int ans = dfs(g, c, u);
mx = max(mx, ans);
}
}
// Add the maximum of
// all children to sub[u]
sub[u] += mx;
// Return maximum sum of
// node 'u' to its parent
return sub[u];
}
// Driver code
int main()
{
vector<int> g[100005];
int edges = 6;
g[18].push_back(7);
g[7].push_back(18);
g[18].push_back(15);
g[15].push_back(18);
g[15].push_back(2);
g[2].push_back(15);
g[7].push_back(4);
g[4].push_back(7);
g[7].push_back(12);
g[12].push_back(7);
g[12].push_back(9);
g[9].push_back(12);
int root = 18;
cout << dfs(g, root, -1);
}
Java
// Java program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
import java.util.Vector;
class GFG{
// Array to store the
// result at each node
static int []sub = new int[100005];
// Function to get minimum difference
// between the divisors of a number
static int minDivisorDifference(int n)
{
int num1 = 0;
int num2 = 0;
// Iterate from square
// root of N to N
for (int i = (int) Math.sqrt(n);
i <= n; i++)
{
if (n % i == 0)
{
num1 = i;
num2 = n / i;
break;
}
}
// return absolute difference
return Math.abs(num1 - num2);
}
// DFS function to calculate
// the maximum sum
static int dfs(Vector<Integer> g[],
int u, int par)
{
// Store the min difference
sub[u] = minDivisorDifference(u);
int mx = 0;
for (int c : g[u])
{
if (c != par)
{
int ans = dfs(g, c, u);
mx = Math.max(mx, ans);
}
}
// Add the maximum of
// all children to sub[u]
sub[u] += mx;
// Return maximum sum of
// node 'u' to its parent
return sub[u];
}
// Driver code
public static void main(String[] args)
{
@SuppressWarnings("unchecked")
Vector<Integer> []g =
new Vector[100005];
for (int i = 0; i < g.length; i++)
g[i] = new Vector<Integer>();
int edges = 6;
g[18].add(7);
g[7].add(18);
g[18].add(15);
g[15].add(18);
g[15].add(2);
g[2].add(15);
g[7].add(4);
g[4].add(7);
g[7].add(12);
g[12].add(7);
g[12].add(9);
g[9].add(12);
int root = 18;
System.out.print(dfs(g, root, -1));
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program to maximize the sum # of minimum difference of divisors # of nodes in an n-ary tree import math # Array to store the # result at each node sub = [0 for i in range(100005)] # Function to get minimum difference # between the divisors of a number def minDivisorDifference(n): num1 = 0 num2 = 0 # Iterate from square # root of N to N for i in range(int(math.sqrt(n)), n + 1): if (n % i == 0): num1 = i num2 = n // i break # Return absolute difference return abs(num1 - num2) # DFS function to calculate the maximum sum def dfs(g, u, par): # Store the min difference sub[u] = minDivisorDifference(u) mx = 0 for c in g[u]: if (c != par): ans = dfs(g, c, u) mx = max(mx, ans) # Add the maximum of # all children to sub[u] sub[u] += mx # Return maximum sum of # node 'u' to its parent return sub[u] # Driver code if __name__=='__main__': g = [[] for i in range(100005)] edges = 6 g[18].append(7) g[7].append(18) g[18].append(15) g[15].append(18) g[15].append(2) g[2].append(15) g[7].append(4) g[4].append(7) g[7].append(12) g[12].append(7) g[12].append(9) g[9].append(12) root = 18 print(dfs(g, root, -1)) # This code is contributed by rutvik_56
C#
// C# program to maximize the sum of
// minimum difference of divisors
// of nodes in an n-ary tree
using System;
using System.Collections.Generic;
class GFG{
// Array to store the
// result at each node
static int []sub = new int[100005];
// Function to get minimum difference
// between the divisors of a number
static int minDivisorDifference(int n)
{
int num1 = 0;
int num2 = 0;
// Iterate from square
// root of N to N
for (int i = (int) Math.Sqrt(n);
i <= n; i++)
{
if (n % i == 0)
{
num1 = i;
num2 = n / i;
break;
}
}
// return absolute difference
return Math.Abs(num1 - num2);
}
// DFS function to calculate
// the maximum sum
static int dfs(List<int> []g,
int u, int par)
{
// Store the min difference
sub[u] = minDivisorDifference(u);
int mx = 0;
foreach (int c in g[u])
{
if (c != par)
{
int ans = dfs(g, c, u);
mx = Math.Max(mx, ans);
}
}
// Add the maximum of
// all children to sub[u]
sub[u] += mx;
// Return maximum sum of
// node 'u' to its parent
return sub[u];
}
// Driver code
public static void Main(String[] args)
{
List<int> []g =
new List<int>[100005];
for (int i = 0; i < g.Length; i++)
g[i] = new List<int>();
int edges = 6;
g[18].Add(7);
g[7].Add(18);
g[18].Add(15);
g[15].Add(18);
g[15].Add(2);
g[2].Add(15);
g[7].Add(4);
g[4].Add(7);
g[7].Add(12);
g[12].Add(7);
g[12].Add(9);
g[9].Add(12);
int root = 18;
Console.Write(dfs(g, root, -1));
}
}
// This code is contributed by Rajput-Ji
Javascript
<script>
// Javascript program to maximize the sum of
// minimum difference of divisors
// Array to store the
// result at each node
let sub = new Array(100005);
sub.fill(0);
// Function to get minimum difference
// between the divisors of a number
function minDivisorDifference(n)
{
let num1 = 0;
let num2 = 0;
// Iterate from square
// root of N to N
for (let i = parseInt(Math.sqrt(n), 10); i <= n; i++)
{
if (n % i == 0)
{
num1 = i;
num2 = parseInt(n / i, 10);
break;
}
}
// return absolute difference
return Math.abs(num1 - num2);
}
// DFS function to calculate
// the maximum sum
function dfs(g, u, par)
{
// Store the min difference
sub[u] = minDivisorDifference(u);
let mx = 0;
for (let c = 0; c < g[u].length; c++)
{
if (g[u] != par)
{
let ans = dfs(g, g[u], u);
mx = Math.max(mx, ans);
}
}
// Add the maximum of
// all children to sub[u]
sub[u] += mx;
// Return maximum sum of
// node 'u' to its parent
return sub[u];
}
let g = new Array(100005);
for (let i = 0; i < g.length; i++)
g[i] = [];
let edges = 6;
g[18].push(7);
g[7].push(18);
g[18].push(15);
g[15].push(18);
g[15].push(2);
g[2].push(15);
g[7].push(4);
g[4].push(7);
g[7].push(12);
g[12].push(7);
g[12].push(9);
g[9].push(12);
let root = 18;
document.write(dfs(g, root, -1));
// This code is contributed by mukesh07.
</script>
Producción:
10
Publicación traducida automáticamente
Artículo escrito por VikasVishwakarma1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA