求加权和最小的子树的根
给定一棵树和所有节点的权重，任务是找到加权和最小的子树的根。
示例:
输入:
输出: 5 父 1 的子树权重=(-1)+(5)+(-2)+(-1)+(3))= 4 父 2 的子树权重= ((5) + (-1) + (3)) = 7 父 3 的子树权重= -1 父 4 的子树权重= 3 父 5 的子树权重= -2 节点 5 给出最小的子树权重
方法:对树执行 dfs ，对每个节点计算以当前节点为根的子树加权和，然后找到一个节点的最小和值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
int ans = 0, mini = INT_MAX;
vector<int> graph[100];
vector<int> weight(100);
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
void dfs(int node, int parent)
{
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum
void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++) {
// If current node's weight
// is minimum so far
if (mini > weight[i]) {
mini = weight[i];
ans = i;
}
}
}
// Driver code
int main()
{
int n = 5;
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// 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);
findMin(n);
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 ans = 0, mini = Integer.MAX_VALUE;
@SuppressWarnings("unchecked")
static Vector<Integer>[] graph = new Vector[100];
static Integer[] weight = new Integer[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
for (int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum x
static void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight x
// is minimum so far
if (mini > weight[i])
{
mini = weight[i];
ans = i;
}
}
}
// Driver code
public static void main(String[] args)
{
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new Vector<Integer>();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
dfs(1, 1);
findMin(n);
System.out.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, mini = int.MaxValue;
static List<int>[] graph = new List<int>[100];
static int[] weight = new int[100];
// Function to perform dfs and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
static void dfs(int node, int parent)
{
foreach (int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[to];
}
}
// Function to find the node
// having minimum sub-tree sum x
static void findMin(int n)
{
// For every node
for (int i = 1; i <= n; i++)
{
// If current node's weight x
// is minimum so far
if (mini > weight[i])
{
mini = weight[i];
ans = i;
}
}
}
// Driver code
public static void Main(String[] args)
{
int n = 5;
for (int i = 0; i < 100; i++)
graph[i] = new List<int>();
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
dfs(1, 1);
findMin(n);
Console.Write(ans);
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
ans = 0
mini = 2**32
graph = [[] for i in range(100)]
weight = [0]*100
# Function to perform dfs and update the tree
# such that every node's weight is the sum of
# the weights of all the nodes in the sub-tree
# of the current node including itself
def dfs(node, parent):
global mini, graph, weight, ans
for to in graph[node]:
if (to == parent):
continue
dfs(to, node)
# Calculating the weighted
# sum of the subtree
weight[node] += weight[to]
# Function to find the node
# having minimum sub-tree sum
def findMin(n):
global mini, graph, weight, ans
# For every node
for i in range(1, n + 1):
# If current node's weight
# is minimum so far
if (mini > weight[i]):
mini = weight[i]
ans = i
# Driver code
n = 5
# Weights of the node
weight[1] = -1
weight[2] = 5
weight[3] = -1
weight[4] = 3
weight[5] = -2
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
dfs(1, 1)
findMin(n)
print(ans)
# This code is contributed by SHUBHAMSINGH10
java 描述语言
<script>
// Javascript implementation of the approach
let ans = 0;
let mini = 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 and update the tree
// such that every node's weight is the sum of
// the weights of all the nodes in the sub-tree
// of the current node including itself
function dfs(node, parent)
{
for(let to = 0; to < graph[node].length; to++)
{
if (graph[node][to] == parent)
continue
dfs(graph[node][to], node);
// Calculating the weighted
// sum of the subtree
weight[node] += weight[graph[node][to]];
}
}
// Function to find the node
// having minimum sub-tree sum
function findMin(n)
{
// For every node
for(let i = 1; i <= n; i++)
{
// If current node's weight
// is minimum so far
if (mini > weight[i])
{
mini = weight[i];
ans = i;
}
}
}
// Driver code
let n = 5;
// Weights of the node
weight[1] = -1;
weight[2] = 5;
weight[3] = -1;
weight[4] = 3;
weight[5] = -2;
// Edges of the tree
graph[1].push(2);
graph[2].push(3);
graph[2].push(4);
graph[1].push(5);
dfs(1, 1);
findMin(n);
document.write(ans);
// This code is contributed by Dharanendra L V.
</script>
Output:
5
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)。因此，时间复杂度为 O(N)。
- 辅助空间: O(n)。 递归栈。
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