计数与子树节点连接时构成八卦的树的节点
给定一棵树，以及所有节点的权重（以字符串的形式），任务是对那些节点的权重字符串与子树节点的字符串连接在一起时，成为节点进行计数。
Pangram：Pangram 是包含英语字母表中每个字母的句子。
示例：，
输入：
输出：1 只有节点 1 的子树的加权字符串构成 Pangram。
方法：在树上执行 dfs 并更新每个节点的权重，以使其存储其权重与子树节点的权重关联。 然后，计算其更新后的加权字符串构成一个聚合图的节点。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
vector<int> graph[100];
vector<string> weight(100);
// Function that returns if the
// string x is a pangram
bool Pangram(string x)
{
map<char, int> mp;
int n = x.size();
for (int i = 0; i < n; i++)
mp[x[i]]++;
if (mp.size() == 26)
return true;
else
return false;
}
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
int countTotalPangram(int n)
{
int cnt = 0;
for (int i = 1; i <= n; i++)
if (Pangram(weight[i]))
cnt++;
return cnt;
}
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
void dfs(int node, int parent)
{
for (int to : graph[node]) {
if (to == parent)
continue;
dfs(to, node);
weight[node] += weight[to];
}
}
// Driver code
int main()
{
int n = 6;
// Weights of the nodes
weight[1] = "abcde";
weight[2] = "fghijkl";
weight[3] = "abcdefg";
weight[4] = "mnopqr";
weight[5] = "stuvwxy";
weight[6] = "zabcdef";
// Edges of the tree
graph[1].push_back(2);
graph[2].push_back(3);
graph[2].push_back(4);
graph[1].push_back(5);
graph[5].push_back(6);
dfs(1, 1);
cout << countTotalPangram(n);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG{
@SuppressWarnings("unchecked")
static Vector<Integer> []graph = new Vector[100];
static String []weight = new String[100];
// Function that returns if the
// String x is a pangram
static boolean Pangram(String x)
{
HashMap<Character, Integer> mp = new HashMap<>();
int n = x.length();
for(int i = 0 ; i < n; i++)
{
if (mp.containsKey(x.charAt(i)))
{
mp.put(x.charAt(i),
mp.get(x.charAt(i)) + 1);
}
else
{
mp.put(x.charAt(i), 1);
}
}
if (mp.size() == 26)
return true;
else
return false;
}
// Function to return the count of nodes
// which make pangram with the
// sub-tree nodes
static int countTotalPangram(int n)
{
int cnt = 0;
for(int i = 1; i <= n; i++)
if (Pangram(weight[i]))
cnt++;
return cnt;
}
// Function to perform dfs and update the nodes
// such that weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
static void dfs(int node, int parent)
{
for(int to : graph[node])
{
if (to == parent)
continue;
dfs(to, node);
weight[node] += weight[to];
}
}
// Driver code
public static void main(String[] args)
{
int n = 6;
// Weights of the nodes
weight[1] = "abcde";
weight[2] = "fghijkl";
weight[3] = "abcdefg";
weight[4] = "mnopqr";
weight[5] = "stuvwxy";
weight[6] = "zabcdef";
for(int i = 0; i < graph.length; i++)
graph[i] = new Vector<Integer>();
// Edges of the tree
graph[1].add(2);
graph[2].add(3);
graph[2].add(4);
graph[1].add(5);
graph[5].add(6);
dfs(1, 1);
System.out.print(countTotalPangram(n));
}
}
// This code is contributed by Amit Katiyar
Python
# Python3 implementation of the approach
graph = [[] for i in range(100)]
weight = [0] * 100
# Function that returns if the
# string x is a pangram
def Pangram(x):
mp = {}
n = len(x)
for i in range(n):
if x[i] not in mp:
mp[x[i]] = 0
mp[x[i]] += 1
if (len(mp)== 26):
return True
else:
return False
# Function to return the count of nodes
# which make pangram with the
# sub-tree nodes
def countTotalPangram(n):
cnt = 0
for i in range(1, n + 1):
if (Pangram(weight[i])):
cnt += 1
return cnt
# Function to perform dfs and update the nodes
# such that weight[i] will store the weight[i]
# concatenated with the weights of
# all the nodes in the sub-tree
def dfs(node, parent):
for to in graph[node]:
if (to == parent):
continue
dfs(to, node)
weight[node] += weight[to]
# Driver code
n = 6
# Weights of the nodes
weight[1] = "abcde"
weight[2] = "fghijkl"
weight[3] = "abcdefg"
weight[4] = "mnopqr"
weight[5] = "stuvwxy"
weight[6] = "zabcdef"
# Edges of the tree
graph[1].append(2)
graph[2].append(3)
graph[2].append(4)
graph[1].append(5)
graph[5].append(6)
dfs(1, 1)
print(countTotalPangram(n))
# This code is contributed by SHUBHAMSINGH10
C
// C# implementation of
// the above approach
using System;
using System.Collections.Generic;
class GFG{
static List<int> []graph =
new List<int>[100];
static String []weight =
new String[100];
// Function that returns if the
// String x is a pangram
static bool Pangram(String x)
{
Dictionary<char,
int> mp = new Dictionary<char,
int>();
int n = x.Length;
for(int i = 0 ; i < n; i++)
{
if (mp.ContainsKey(x[i]))
{
mp[x[i]] = mp[x[i]] + 1;
}
else
{
mp.Add(x[i], 1);
}
}
if (mp.Count == 26)
return true;
else
return false;
}
// Function to return the
// count of nodes which
// make pangram with the
// sub-tree nodes
static int countTotalPangram(int n)
{
int cnt = 0;
for(int i = 1; i <= n; i++)
if (Pangram(weight[i]))
cnt++;
return cnt;
}
// Function to perform dfs and
// update the nodes such that
// weight[i] will store the weight[i]
// concatenated with the weights of
// all the nodes in the sub-tree
static void dfs(int node, int parent)
{
foreach(int to in graph[node])
{
if (to == parent)
continue;
dfs(to, node);
weight[node] += weight[to];
}
}
// Driver code
public static void Main(String[] args)
{
int n = 6;
// Weights of the nodes
weight[1] = "abcde";
weight[2] = "fghijkl";
weight[3] = "abcdefg";
weight[4] = "mnopqr";
weight[5] = "stuvwxy";
weight[6] = "zabcdef";
for(int i = 0;
i < graph.Length; i++)
graph[i] = new List<int>();
// Edges of the tree
graph[1].Add(2);
graph[2].Add(3);
graph[2].Add(4);
graph[1].Add(5);
graph[5].Add(6);
dfs(1, 1);
Console.Write(countTotalPangram(n));
}
}
// This code is contributed by shikhasingrajput
Output:
1
复杂度分析：
-
时间复杂度：O（N * S）。
在 dfs 中，树的每个节点都处理一次，因此，如果树中总共有 N 个节点，则由于 dfs 而导致的复杂度为
O(N)。 同样，为了处理每个节点，将 Pangram（）函数用于复杂度为 O（S）的每个节点，其中 S 是子树中所有权重字符串的长度之和，并且由于对每个节点都执行了此操作，因此 这部分的总时间复杂度变为 O（N * S）。 因此，最终时间复杂度为 O（N * S）。 -
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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