给定一个由N 个节点组成的N叉树,其值为0到(N – 1) ,任务是找到给定树中存在的子树总数。由于计数可能非常大,所以打印计数模 1000000007。
例子:
Input: N = 3
0
/
1
/
2
Output: 7
Explanation:
The total number of subtrees nodes are {}, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 1, 2}.
Input: N = 2
0
/
1
Output: 4
方法:解决给定问题的方法是对给定树执行 DFS 遍历。请按照以下步骤解决问题:
- 初始化一个变量,比如count为0 ,以存储给定树中存在的子树总数的计数。
- 声明一个函数DFS(int src, int parent)来统计节点src的子树数,并执行以下操作:
- 初始化一个变量,比如res为1 。
- 遍历当前节点的邻接表,如果邻接表中的节点,说X与父节点不一样,则递归调用节点X的DFS函数和作为父节点的节点src为DFS(X,源) 。
- 让上述递归调用返回的值为 value ,然后将res的值更新为(res * (value + 1)) % (10 9 + 7) 。
- 将count的值更新为(count + res) % (10 9 + 7) 。
- 从每个递归调用返回res的值。
- 为根节点1调用函数DFS() 。
- 完成以上步骤后,打印count的值作为结果。
下面是上述方法的实现:
C++
// C++ program of the above approach
#include
#define MAX 300004
using namespace std;
// Adjacency list to
// represent the graph
vector graph[MAX];
int mod = 1e9 + 7;
// Stores the count of subtrees
// possible from given N-ary Tree
int ans = 0;
// Utility function to count the number of
// subtrees possible from given N-ary Tree
int countSubtreesUtil(int cur, int par)
{
// Stores the count of subtrees
// when cur node is the root
int res = 1;
// Traverse the adjacency list
for (int i = 0;
i < graph[cur].size(); i++) {
// Iterate over every ancestor
int v = graph[cur][i];
if (v == par)
continue;
// Calculate product of the number
// of subtrees for each child node
res = (res
* (countSubtreesUtil(
v, cur)
+ 1))
% mod;
}
// Update the value of ans
ans = (ans + res) % mod;
// Return the resultant count
return res;
}
// Function to count the number of
// subtrees in the given tree
void countSubtrees(int N,
vector >& adj)
{
// Initialize an adjacency matrix
for (int i = 0; i < N - 1; i++) {
int a = adj[i].first;
int b = adj[i].second;
// Add the edges
graph[a].push_back(b);
graph[b].push_back(a);
}
// Function Call to count the
// number of subtrees possible
countSubtreesUtil(1, 1);
// Print count of subtrees
cout << ans + 1;
}
// Driver Code
int main()
{
int N = 3;
vector > adj
= { { 0, 1 }, { 1, 2 } };
countSubtrees(N, adj);
return 0;
}
Java
// Java program of above approach
import java.util.*;
class GFG{
static int MAX = 300004;
// Adjacency list to
// represent the graph
static ArrayList> graph;
static long mod = (long)1e9 + 7;
// Stores the count of subtrees
// possible from given N-ary Tree
static int ans = 0;
// Utility function to count the number of
// subtrees possible from given N-ary Tree
static int countSubtreesUtil(int cur, int par)
{
// Stores the count of subtrees
// when cur node is the root
int res = 1;
// Traverse the adjacency list
for(int i = 0;
i < graph.get(cur).size(); i++)
{
// Iterate over every ancestor
int v = graph.get(cur).get(i);
if (v == par)
continue;
// Calculate product of the number
// of subtrees for each child node
res = (int)((res * (countSubtreesUtil(
v, cur) + 1)) % mod);
}
// Update the value of ans
ans = (int)((ans + res) % mod);
// Return the resultant count
return res;
}
// Function to count the number of
// subtrees in the given tree
static void countSubtrees(int N, int[][] adj)
{
// Initialize an adjacency matrix
for(int i = 0; i < N - 1; i++)
{
int a = adj[i][0];
int b = adj[i][1];
// Add the edges
graph.get(a).add(b);
graph.get(b).add(a);
}
// Function Call to count the
// number of subtrees possible
countSubtreesUtil(1, 1);
// Print count of subtrees
System.out.println(ans + 1);
}
// Driver code
public static void main(String[] args)
{
int N = 3;
int[][] adj = { { 0, 1 }, { 1, 2 } };
graph = new ArrayList<>();
for(int i = 0; i < MAX; i++)
graph.add(new ArrayList<>());
countSubtrees(N, adj);
}
}
// This code is contributed by offbeat
Python3
# Python3 program of the above approach
MAX = 300004
# Adjacency list to
# represent the graph
graph = [[] for i in range(MAX)]
mod = 10**9 + 7
# Stores the count of subtrees
# possible from given N-ary Tree
ans = 0
# Utility function to count the number of
# subtrees possible from given N-ary Tree
def countSubtreesUtil(cur, par):
global mod, ans
# Stores the count of subtrees
# when cur node is the root
res = 1
# Traverse the adjacency list
for i in range(len(graph[cur])):
# Iterate over every ancestor
v = graph[cur][i]
if (v == par):
continue
# Calculate product of the number
# of subtrees for each child node
res = (res * (countSubtreesUtil(v, cur)+ 1)) % mod
# Update the value of ans
ans = (ans + res) % mod
# Return the resultant count
return res
# Function to count the number of
# subtrees in the given tree
def countSubtrees(N, adj):
# Initialize an adjacency matrix
for i in range(N-1):
a = adj[i][0]
b = adj[i][1]
# Add the edges
graph[a].append(b)
graph[b].append(a)
# Function Call to count the
# number of subtrees possible
countSubtreesUtil(1, 1)
# Prcount of subtrees
print (ans + 1)
# Driver Code
if __name__ == '__main__':
N = 3
adj = [ [ 0, 1 ], [ 1, 2 ] ]
countSubtrees(N, adj)
# This code is contributed by mohit kumar 29.
C#
// C# program of above approach
using System;
using System.Collections.Generic;
public class GFG
{
static int MAX = 300004;
// Adjacency list to
// represent the graph
static List> graph;
static long mod = (long) 1e9 + 7;
// Stores the count of subtrees
// possible from given N-ary Tree
static int ans = 0;
// Utility function to count the number of
// subtrees possible from given N-ary Tree
static int countSubtreesUtil(int cur, int par) {
// Stores the count of subtrees
// when cur node is the root
int res = 1;
// Traverse the adjacency list
for (int i = 0; i < graph[cur].Count; i++) {
// Iterate over every ancestor
int v = graph[cur][i];
if (v == par)
continue;
// Calculate product of the number
// of subtrees for each child node
res = (int) ((res * (countSubtreesUtil(v, cur) + 1)) % mod);
}
// Update the value of ans
ans = (int) ((ans + res) % mod);
// Return the resultant count
return res;
}
// Function to count the number of
// subtrees in the given tree
static void countSubtrees(int N, int[,] adj) {
// Initialize an adjacency matrix
for (int i = 0; i < N - 1; i++) {
int a = adj[i,0];
int b = adj[i,1];
// Add the edges
graph[a].Add(b);
graph[b].Add(a);
}
// Function Call to count the
// number of subtrees possible
countSubtreesUtil(1, 1);
// Print count of subtrees
Console.WriteLine(ans + 1);
}
// Driver code
public static void Main(String[] args) {
int N = 3;
int[,] adj = { { 0, 1 }, { 1, 2 } };
graph = new List>();
for (int i = 0; i < MAX; i++)
graph.Add(new List());
countSubtrees(N, adj);
}
}
// This code is contributed by Amit Katiyar
Javascript
输出:
7
时间复杂度: O(N)
辅助空间: O(N)
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