找到与 X 的和具有最小设置位的节点
原文:https://www . geeksforgeeks . org/find-the-node-with-x-sum-with-x-with-minimum-set-bits/
给定一棵树，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 给出最小的设置位，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。
示例:
输入:
x = 15 输出: 1 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 具有最小设置位的节点。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。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int minimum = INT_MAX, x, ans = INT_MAX;
vector<int> graph[100];
vector<int> weight(100);
// Function to perform dfs to find
// the minimum set bits value
void dfs(int node, int parent)
{
// If current set bits value is smaller than
// the current minimum
int a = __builtin_popcount(weight[node] + x);
if (minimum > a) {
minimum = a;
ans = node;
}
// If count is equal to the minimum
// then choose the node with minimum value
else if (minimum == 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 minimum = Integer.MAX_VALUE,
x, ans = Integer.MAX_VALUE;
static Vector<Integer> []graph =
new Vector[100];
static int []weight = new int[100];
// Function to perform dfs
// to find the minimum set
// bits value
static void dfs(int node,
int parent)
{
// If current set bits value
// is smaller than the current
// minimum
int a = Integer.bitCount(weight[node] + x);
if (minimum > a)
{
minimum = a;
ans = node;
}
// If count is equal to the
// minimum then choose the
// node with minimum value
else if (minimum == a)
ans = Math.min(ans, node);
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver code
public static void main(String[] args)
{
x = 15;
for (int i = 0; i < graph.length; i++)
graph[i] = new Vector<Integer>();
// 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].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
System.out.print(ans);
}
}
// This code is contributed by gauravrajput1
Python 3
# Python3 implementation of the approach
from sys import maxsize
minimum, x, ans = maxsize, None, maxsize
graph = [[] for i in range(100)]
weight = [0] * 100
# Function to perform dfs to find
# the minimum set bits value
def dfs(node, parent):
global x, ans, graph, weight, minimum
# If current set bits value is greater than
# the current minimum
a = bin(weight[node] + x).count('1')
if minimum > a:
minimum = a
ans = node
# If count is equal to the minimum
# then choose the node with minimum value
elif minimum == 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;
using System.Collections.Generic;
using System.Text;
class GFG{
static int minimum = int.MaxValue, x,
ans = int.MaxValue;
static ArrayList[] graph = new ArrayList[100];
static int[] weight = new int[100];
static int PopCount(int n)
{
int count = 0;
while (n > 0)
{
count += n & 1;
n >>= 1;
}
return count;
}
// Function to perform dfs to find
// the minimum set bits value
static void dfs(int node, int parent)
{
// If current set bits value is smaller
// than the current minimum
int a = PopCount(weight[node] + x);
if (minimum > a)
{
minimum = a;
ans = node;
}
// If count is equal to the minimum
// then choose the node with minimum value
else if (minimum == a)
ans = Math.Min(ans, node);
foreach(int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
}
}
// Driver Code
public static void Main(string[] args)
{
x = 15;
for(int i = 0; i < 100; i++)
graph[i] = new ArrayList();
// 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].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 rutvik_56
java 描述语言
<script>
// Javascript implementation of the approach
let minimum = Number.MAX_VALUE;
let x;
let ans = Number.MAX_VALUE;
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 to find
// the minimum set bits value
function dfs(node, parent)
{
// If current set bits value is smaller than
// the current minimum
let a = (weight[node] + x).toString(2).split('').filter(
y => y == '1').length;
if (minimum > a)
{
minimum = a;
ans = node;
}
// If count is equal to the minimum
// then choose the node with minimum value
else if (minimum == a)
ans = Math.min(ans, node);
for(let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue
dfs(graph[node][to], node);
}
}
// Driver code
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(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
document.write(ans);
// This code is contributed by Dharanendra L V.
</script>
Output:
1
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，由于该复杂度是常数，因此它不会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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