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:
Input:
x = 15
Output: 4
Node 1: setbits(5 + 15) = 2
Node 2: setbits(10 + 15) = 3
Node 3: setbits(11 + 15) = 3
Node 4: setbits(8 + 15) = 4
Node 5: setbits(6 + 15) = 3
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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