Dado un árbol y los pesos de todos los Nodes, la tarea es encontrar la raíz del subárbol cuya suma ponderada XOR con el entero X dado es máxima.
Ejemplos:
Aporte:
X = 15
Salida: 4
Peso del subárbol para padre 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11
Peso del sub -árbol para padre 2 = ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8
Peso del subárbol para padre 3 = -1 XOR 15 = -16
Peso del subárbol para padre 4 = 3 XOR 15 = 12
Peso del subárbol para el padre 5 = -2 XOR 15 = -15
El Node 4 proporciona la suma máxima ponderada del subárbol XOR X.
Enfoque: Realice dfs en el árbol y, para cada Node, calcule la suma ponderada del subárbol con raíz en el Node actual y luego encuentre el valor máximo (suma XOR X) para un Node.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int ans = 0, maxi = INT_MIN;
vector<int> graph[100];
vector<int> weight(100);
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
void dfs(int node, int parent)
{
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having maximum sub-tree sum XOR x
void findMaxX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++) {
// If current node's weight XOR x
// is maximum so far
if (maxi < (weight[i] ^ x)) {
maxi = (weight[i] ^ x);
ans = i;
}
}
}
// Driver code
int main()
{
int x = 15;
int n = 5;
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// 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);
findMaxX(n, x);
cout << ans;
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
static int ans = 0, maxi = Integer.MIN_VALUE;
static Vector<Integer>[] graph = new Vector[100];
static Integer[] weight = new Integer[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having maximum sub-tree sum XOR x
static void findMaxX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is maximum so far
if (maxi < (weight[i] ^ x))
{
maxi = (weight[i] ^ x);
ans = i;
}
}
}
// Driver code
public static void main(String[] args)
{
int x = 15;
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new Vector<Integer>();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
findMaxX(n, x);
System.out.print(ans);
}
}
// This code is contributed by Rajput-Ji
Python
# Python implementation of the approach from sys import maxsize # Function to perform dfs and update the tree # such that every node's weight is the sum of # the weights of all the nodes in the sub-tree # of the current node including itself def dfs(node, parent): global maxi, graph, weight, x, ans for to in graph[node]: if (to == parent): continue dfs(to, node) # Calculating the weighted # sum of the subtree weight[node] += weight[to] # Function to find the node # having maximum sub-tree sum XOR x def findMaxX(n, x): global maxi, graph, weight, ans # For every node for i in range(1, n + 1): # If current node's weight XOR x # is maximum so far if (maxi < (weight[i] ^ x)): maxi = (weight[i] ^ x) ans = i # Driver code ans = 0 maxi = -maxsize graph = [[] for i in range(100)] weight = [0]*100 x = 15 n = 5 # Weights of the node weight[1] = -1 weight[2] = 5 weight[3] = -1 weight[4] = 3 weight[5] = -2 # Edges of the tree graph[1].append(2) graph[2].append(3) graph[2].append(4) graph[1].append(5) dfs(1, 1) findMaxX(n, x) print(ans) # This code is contributed by SHUBHAMSINGH10
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
static int ans = 0, maxi = int.MinValue;
static List<int>[] graph = new List<int>[100];
static int[] weight = new int[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
foreach (int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having maximum sub-tree sum XOR x
static void findMaxX(int n, int x)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is maximum so far
if (maxi < (weight[i] ^ x))
{
maxi = (weight[i] ^ x);
ans = i;
}
}
}
// Driver code
public static void Main(String[] args)
{
int x = 15;
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new List<int>();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
findMaxX(n, x);
Console.Write(ans);
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// Javascript implementation of the approach
let maxi = Number.MIN_VALUE, x, ans;
let graph = new Array(100);
let weight = new Array(100);
for(let i = 0; i < 100; i++)
{
graph[i] = [];
weight[i] = 0;
}
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
function dfs(node, parent)
{
for(let to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having maximum sub-tree sum XOR x
function findMaxX(n, x)
{
// For every node
for (let i = 1; i <= n; i++)
{
// If current node's weight XOR x
// is maximum so far
if (maxi < (weight[i] ^ x))
{
maxi = (weight[i] ^ x);
ans = i;
}
}
}
// Driver Code
x = 15;
let n = 5;
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
findMaxX(n, x);
document.write(ans);
// This code is contributed by unknown2108
</script>
Producción:
4
Análisis de Complejidad:
- Complejidad temporal: O(N).
En dfs, cada Node del árbol se procesa una vez y, por lo tanto, la complejidad debida a dfs es O(N) si hay un total de N Nodes en el árbol. Por lo tanto, la complejidad del tiempo es O(N). - Espacio Auxiliar : O(n).
Pila de recursividad.
Publicación traducida automáticamente
Artículo escrito por mohit kumar 29 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA