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📜  找到与 X 之和具有最大设置位的节点

📅  最后修改于: 2021-10-25 05:00:58             🧑  作者: Mango

给定一棵树,以及所有节点的权重和整数x ,任务是找到一个节点i使得weight[i] + x具有最大设置位。如果两个或多个节点在添加x时具有相同的设置位计数,则找到具有最小值的节点。
例子:

方法:在树上执行 dfs 并跟踪与x之和具有最大设置位的节点。如果两个或更多节点的设置位数相等,则选择数量最少的一个。
下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
int maximum = INT_MIN, x, ans = INT_MAX;
 
vector graph[100];
vector weight(100);
 
// Function to perform dfs to find
// the maximum set bits value
void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight[node] + x);
    if (maximum < a) {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = min(ans, node);
 
    for (int to : graph[node]) {
        if (to == parent)
            continue;
        dfs(to, node);
    }
}
 
// Driver code
int main()
{
    x = 15;
 
    // 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 maximum = Integer.MIN_VALUE, x, ans = Integer.MAX_VALUE;
 
static Vector> graph = new Vector>();
static Vector weight = new Vector();
 
//number of set bits
static int __builtin_popcount(int x)
{
    int c = 0;
    for(int i = 0; i < 60; i++)
    if(((x>>i)&1) != 0)c++;
     
    return c;
}
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight.get(node) + x);
    if (maximum < a)
    {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = Math.min(ans, node);
         
    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[])
{
    x = 15;
 
    // 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
# Python implementation of the approach
from sys import maxsize
 
maximum, x, ans = -maxsize, None, maxsize
 
graph = [[] for i in range(100)]
weight = [0] * 100
 
# Function to perform dfs to find
# the maximum set bits value
def dfs(node, parent):
    global x, ans, graph, weight, maximum
 
    # If current set bits value is greater than
    # the current maximum
    a = bin(weight[node] + x).count('1')
 
    if maximum < a:
        maximum = a
        ans = node
 
    # If count is equal to the maximum
    # then choose the node with minimum value
    elif maximum == a:
        ans = min(ans, node)
 
    for to in graph[node]:
        if to == parent:
            continue
        dfs(to, node)
 
# Driver Code
if __name__ == "__main__":
 
    x = 15
 
    # 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
# sanjeev2552


C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
static int maximum = int.MinValue, x,ans = int.MaxValue;
 
static List> graph = new List>();
static List weight = new List();
 
// number of set bits
static int __builtin_popcount(int x)
{
    int c = 0;
    for(int i = 0; i < 60; i++)
    if(((x>>i)&1) != 0)c++;
     
    return c;
}
 
// Function to perform dfs to find
// the maximum value
static void dfs(int node, int parent)
{
    // If current set bits value is greater than
    // the current maximum
    int a = __builtin_popcount(weight[node] + x);
    if (maximum < a)
    {
        maximum = a;
        ans = node;
    }
 
    // If count is equal to the maximum
    // then choose the node with minimum value
    else if (maximum == a)
        ans = Math.Min(ans, node);
         
    for (int i = 0; i < graph[node].Count; i++)
    {
        if (graph[node][i] == parent)
            continue;
        dfs(graph[node][i], node);
    }
}
 
// Driver code
public static void Main()
{
    x = 15;
 
    // 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.Write( ans);
}
}
 
// This code is contributed by mits


Javascript


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)。此外,为了处理每个节点,使用了 builtin_popcount()函数,它的复杂度为 O(c),其中 c 是一个常数,并且由于这个复杂度是常数,它不会影响整体时间复杂度。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。

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