求与 X 加权和异或最大的子树的根
原文:https://www . geeksforgeeks . org/find-子树的根-其加权和-xor-with-x-是最大值/
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 加权和异或最大的子树的根。 例:
输入:
X = 15 输出: 4 父 1 的子树权重=(-1)+(5)+(-2)+(-1)+(3))XOR 15 = 4 XOR 15 = 11 父 2 的子树权重= ((5) + (-1) + (3)) XOR 15 = 7 XOR 15 = 8 父 3 的子树权重= -1 XOR 15 = -16 父 4 的子树权重= 3 XOR 15 = 12 父 5 的子树权重= -2 XOR 15 = -15 节点 4 给出最大子树加权和 XOR X.
方法:对树执行 dfs ，对于每个节点，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大值(和异或 X)。 以下是上述方法的实施:
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 语言(一种计算机语言，尤用于创建网站)
// 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
计算机编程语言
# 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
java 描述语言
<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>
Output:
4
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，由于 dfs 而导致的复杂性是 O(N)。因此，时间复杂度为 O(N)。
- 辅助空间: O(n)。 递归栈。
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