查找其与X的加权和XOR最大的子树的根

给定一棵树，以及所有节点的权重，任务是找到其给定整数X的加权和XOR最大的子树的根。
例子：

Input: 
 

X = 15 
Output: 4 
Weight of sub-tree for parent 1 = ((-1) + (5) + (-2) + (-1) + (3)) XOR 15 = 4 XOR 15 = 11 
Weight of sub-tree for parent 2 = ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8 
Weight of sub-tree for parent 3 = -1 XOR 15 = -16 
Weight of sub-tree for parent 4 = 3 XOR 15 = 12 
Weight of sub-tree for parent 5 = -2 XOR 15 = -15 
Node 4 gives the maximum sub-tree weighted sum XOR X. 
 

方法：在树上执行dfs，并为每个节点计算以当前节点为根的子树加权总和，然后找到该节点的最大值(总和XOR X)。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int ans = 0, maxi = INT_MIN;
 
vector graph[100];
vector 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[] 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();
         
        // 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[] graph = new List[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();
         
        // 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

// 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

输出：

4

复杂度分析：

• 时间复杂度： O(N)。
  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)。因此，时间复杂度为O(N)。
• 辅助空间： O(n)。
  递归堆栈。