dsal 找到与 X 的和具有最大设置位的节点

找到与 X 的和具有最大设置位的节点

原文:https://www . geeksforgeeks . org/find-the-node-what-sum-with-x-has-maximum-set-bits/

给定一棵树以及所有节点的权重和一个整数 x ，任务是找到一个节点 i ，使得权重[i] + x 具有最大的设置位。如果两个或多个节点在加上 x 时具有相同的设置位计数，则找到具有最小值的节点。 举例:

输入:

x = 15 输出: 4 节点 1:设置位(5 + 15) = 2 节点 2:设置位(10 + 15) = 3 节点 3:设置位(11 + 15) = 3 节点 4:设置位(8 + 15) = 4 节点 5:设置位(6 + 15) = 3

方法:在树上执行 dfs 并跟踪其与的和 x 具有最大设置位的节点。如果两个或两个以上的节点具有相同数量的设置位，则选择具有最小数量的节点。以下是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

int maximum = INT_MIN, x, ans = INT_MAX;

vector<int> graph[100];
vector<int> 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 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 maximum = Integer.MIN_VALUE, x, ans = Integer.MAX_VALUE;

static Vector<Vector<Integer>> graph = new Vector<Vector<Integer>>();
static Vector<Integer> weight = new Vector<Integer>();

//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<Integer>());

    // 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

Python 3

# 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<List<int>> graph = new List<List<int>>();
static List<int> weight = new List<int>();

// 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<int>());

    // 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

java 描述语言

<script>

// Javascript implementation of the approach

    let maximum = Number.MIN_VALUE, x,
    ans = Number.MAX_VALUE;

    let graph = new Array(100);
    for(let i=0;i<100;i++)
    {
        graph[i]=[];
    }

    let weight = [];

    //number of set bits
    function __builtin_popcount(x)
    {
        let c = 0;
        for(let i = 0; i < 60; i++)
            if(((x>>i)&1) != 0)
                c++;

        return c;
    }

    // Function to perform dfs to find
   // the maximum value
    function dfs(node,parent)
    {
        // If current set bits value is greater than
        // the current maximum
        let 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 (let i = 0; i < graph[node].length; i++)
        {
            if (graph[node][i] == parent)
                continue;
            dfs(graph[node][i], node);
        }
    }

    // Driver code

    x = 15;

    // Weights of the node
    weight.push(0);
    weight.push(5);
    weight.push(10);;
    weight.push(11);;
    weight.push(8);
    weight.push(6);

    // Edges of the tree
    graph[1].push(2);
    graph[2].push(3);
    graph[2].push(4);
    graph[1].push(5);

    dfs(1, 1);

    document.write( ans);

    // This code is contributed by unknown2108

</script>

Output:

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)。此外，为了处理每个节点，使用了 builtin_popcount()函数，该函数的复杂度为 O(c)，其中 c 是常数，并且由于该复杂度是常数，因此它不影响总的时间复杂度。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，所以空间复杂度不变。

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找到与 X 的和具有最大设置位的节点

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dsal 找到与 X 的和具有最大设置位的节点

找到与 X 的和具有最大设置位的节点

C++

Java 语言(一种计算机语言，尤用于创建网站)

Python 3

C

java 描述语言

留言

联系客服

dsal 查看更多书籍