求与X的加权和异或最小的子树的根

给定一棵树，以及所有节点的权重，任务是找到与给定整数X的加权和 XOR 最小的子树的根。
例子：

Input: 
 

X = 15 
Output: 5 
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 3 gives the minimum sub-tree weighted sum XOR X. 
 

方法：对树进行dfs，对每个节点计算以当前节点为根的子树加权和，然后找到一个节点的最小值（sum XOR X）值。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
int ans = 0, mini = INT_MAX;
 
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 minimum sub-tree sum XOR x
void findMinX(int n, int x)
{
 
    // For every node
    for (int i = 1; i <= n; i++) {
 
        // If current node's weight XOR x
        // is minimum so far
        if (mini > (weight[i] ^ x)) {
            mini = (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);
    findMinX(n, x);
 
    cout << ans;
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
    static int ans = 0, mini = Integer.MAX_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 minimum sub-tree sum XOR x
    static void findMinX(int n, int x)
    {
 
        // For every node
        for (int i = 1; i <= n; i++)
        {
 
            // If current node's weight XOR x
            // is minimum so far
            if (mini > (weight[i] ^ x))
            {
                mini = (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);
        findMinX(n, x);
 
        System.out.print(ans);
    }
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
ans = 0
mini = 2**32
 
graph = [[] for i in range(100)]
weight = [0] * 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
def dfs(node, parent):
    global ans, mini, graph, weight, x
    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 minimum sub-tree sum XOR x
def findMinX(n, x):
    global ans, mini,graph,weight
     
    # For every node
    for i in range(1, n + 1):
         
        # If current node's weight XOR x
        # is minimum so far
        if (mini > (weight[i] ^ x)):
            mini = (weight[i] ^ x)
            ans = i
             
# Driver code
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)
findMinX(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, mini = int.MaxValue;
 
    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 minimum sub-tree sum XOR x
    static void findMinX(int n, int x)
    {
 
        // For every node
        for (int i = 1; i <= n; i++)
        {
 
            // If current node's weight XOR x
            // is minimum so far
            if (mini > (weight[i] ^ x))
            {
                mini = (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);
        findMinX(n, x);
 
        Console.Write(ans);
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// Javascript implementation of the approach
     
    let ans = 0, mini = Number.MAX_VALUE;
    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=0;to<graph[node].length;to++)
        {
            if (graph[node][to] == parent)
                continue;
            dfs(graph[node][to], node);
 
            // Calculating the weighted
            // sum of the subtree
            weight[node] += weight[to];
        }
    }
     
    // Function to find the node
    // having minimum sub-tree sum XOR x
    function findMinX(n,X)
    {
        // For every node
        for (let i = 1; i <= n; i++)
        {
 
            // If current node's weight XOR x
            // is minimum so far
            if (mini > (weight[i] ^ x))
            {
                mini = (weight[i] ^ x);
                ans = i;
            }
        }
    }
     
    // Driver code
    let 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);
        findMinX(n, x);
 
        document.write(ans);
 
     
    // This code is contributed by unknown2108
 

输出：

3

复杂度分析：

• 时间复杂度： O(N)。
  在 dfs 中，树的每个节点都被处理一次，因此如果树中总共有 N 个节点，则由于 dfs 的复杂性是 O(N)。因此，时间复杂度为 O(N)。
• 辅助空间： O(n)。
  递归堆栈空间可以达到 O(n)。

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