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📜  计算给定树中权重为偶校验的节点

📅  最后修改于: 2021-05-07 01:46:34             🧑  作者: Mango

给定一棵树和所有节点的权重，任务是计算权重为偶校验的节点数，即它们中的设置位数是否为偶数。

例子：

方法：在树上执行dfs，并为每个节点检查其权重是否相等。如果是，则增加计数。

下面是上述方法的实现：

C++
// C++ implementation of the approach
#include 
using namespace std;
  
int ans = 0;
  
vector graph[100];
vector weight(100);
  
// Function that returns true if count
// of set bits in x is even
bool isEvenParity(int x)
{
    // parity will store the
    // count of set bits
    int parity = 0;
    while (x != 0) {
        x = x & (x - 1);
        parity++;
    }
  
    if (parity % 2 == 0)
        return true;
    else
        return false;
}
  
// Function to perform dfs
void dfs(int node, int parent)
{
    // If weight of the current
    // node has even parity
    if (isEvenParity(weight[node]))
        ans += 1;
  
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
  
// Driver code
int main()
{
    // Weights of the node
    weight[1] = 5;
    weight[2] = 10;
    weight[3] = 11;
    weight[4] = 8;
    weight[5] = 6;
  
    // 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);
  
    cout << ans;
  
    return 0;
}

Java
// Java implementation of the approach 
import java.util.*;
  
class GFG
{
  
static int ans = 0; 
  
static Vector> graph = new Vector>(); 
static Vector weight = new Vector(); 
  
// Function that returns true if count 
// of set bits in x is even 
static boolean isEvenParity(int x) 
{ 
    // parity will store the 
    // count of set bits 
    int parity = 0; 
    while (x != 0) 
    { 
        x = x & (x - 1); 
        parity++; 
    } 
  
    if (parity % 2 == 0) 
        return true; 
    else
        return false; 
} 
  
// Function to perform dfs 
static void dfs(int node, int parent) 
{ 
    // If weight of the current 
    // node has even parity 
    if (isEvenParity(weight.get(node) ))
        ans += 1; 
  
    for (int i = 0; i < graph.get(node).size(); i++) 
    { 
        if (graph.get(node).get(i) == parent) 
            continue; 
        dfs(graph.get(node).get(i) , node); 
    } 
} 
  
// Driver code 
public static void main(String args[])
{ 
    // Weights of the node 
    weight.add( 0); 
    weight.add( 5); 
    weight.add( 10);; 
    weight.add( 11);; 
    weight.add( 8); 
    weight.add( 6); 
  
    for(int i=0;i<100;i++)
    graph.add(new Vector());
      
    // Edges of the tree 
    graph.get(1).add(2); 
    graph.get(2).add(3); 
    graph.get(2).add(4); 
    graph.get(1).add(5); 
  
    dfs(1, 1); 
  
    System.out.println( ans ); 
  
}
}
  
// This code is contributed by Arnab Kundu

Python3
# Python3 implementation of the approach
ans = 0
  
graph = [[] for i in range(100)]
weight = [0]*100
  
# Function that returns True if count
# of set bits in x is even
def isEvenParity(x):
  
    # parity will store the
    # count of set bits
    parity = 0
    while (x != 0):
        x = x & (x - 1)
        parity += 1
          
    if (parity % 2 == 0):
        return True
    else:
        return False
  
# Function to perform dfs
def dfs(node, parent):
    global ans
      
    # If weight of the current
    # node has even parity
    if (isEvenParity(weight[node])):
        ans += 1
      
    for to in graph[node]:
        if (to == parent):
            continue
        dfs(to, node)
  
# Driver code
  
# Weights of the node
weight[1] = 5
weight[2] = 10
weight[3] = 11
weight[4] = 8
weight[5] = 6
  
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
  
dfs(1, 1)
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; 
  
static List> graph = new List>();
static List weight = new List();
  
// Function that returns true if count 
// of set bits in x is even 
static bool isEvenParity(int x) 
{ 
    // parity will store the 
    // count of set bits 
    int parity = 0; 
    while (x != 0) 
    { 
        x = x & (x - 1); 
        parity++; 
    } 
  
    if (parity % 2 == 0) 
        return true; 
    else
        return false; 
} 
  
// Function to perform dfs 
static void dfs(int node, int parent) 
{ 
    // If weight of the current 
    // node has even parity 
    if (isEvenParity(weight[node]))
        ans += 1; 
  
    for (int i = 0; i < graph[node].Count; i++) 
    { 
        if (graph[node][i] == parent) 
            continue; 
        dfs(graph[node][i] , node); 
    } 
} 
  
// Driver code 
static void Main()
{ 
    // Weights of the node 
    weight.Add(0); 
    weight.Add(5); 
    weight.Add(10);
    weight.Add(11);
    weight.Add(8); 
    weight.Add(6); 
  
    for(int i = 0; i < 100; i++)
    graph.Add(new List());
      
    // Edges of the tree 
    graph[1].Add(2); 
    graph[2].Add(3); 
    graph[2].Add(4); 
    graph[1].Add(5); 
  
    dfs(1, 1); 
  
    Console.WriteLine( ans ); 
}
}
  
// This code is contributed by mits

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，因此对于树中的N个节点，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(1)。
    不需要任何额外的空间，因此空间复杂度是恒定的。