解釋：

You are given a data structure of employee information, which includes the employee’s unique id, his importance value and his direct subordinates’ id. For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct. Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates. Example 1:

Input: [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1 
Output:11 
Explanation: 
Employee 1 has importance value 5, and he has two direct subordinates: employee 2
 and employee 3. They both have importance value 3. So the total importance value of 
 employee 1 is 5 + 3 + 3 = 11. 

Note: One employee has at most one direct leader and may have several subordinates. The maximum number of employees won’t exceed 2000.

解釋： 一看題目就是經典的dfs。 python程式碼：

"""
# Employee info
class Employee(object):
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
"""
 

class Solution(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        sum=0
        for employee in employees:
            if id==employee.id:
                sum+=employee.importance
                for subordinate in employee.subordinates:
                    sum+=self.getImportance(employees,subordinate)
        return sum       

/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        //寫個遞迴
        int sum=0;
        for(auto employee:employees)
        {
            if (employee->id==id)
            {   sum+=employee->importance;
                for(auto sub:employee->subordinates)
                    sum+=getImportance(employees,sub);
            }
        }
        return sum; 
    }
};

用字典存起來更快一點（為什麼別人的程式碼快但是我的還是一樣慢？？），dfs最好把字典作為一個形參，這樣用c++實現也更方便一點（nonono，突然發現c++也可以用一個全域性的變數使得兩個函式都可以使用，所以c++也不需要傳太多形式引數，哈哈哈哈），dfs函式的形參也可以是dfs(int id,map<>)這樣似乎更快一點？ python程式碼：

"""
"""
# Employee info
class Employee(object):
    def __init__(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
"""
class Solution(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        
        """
        def dfs(root,_dict):
            if not root:
                return 0
            return root.importance+sum([dfs(_dict[sub],_dict) for sub in root.subordinates])
        _dict={e.id:e for e in employees} 
        return dfs(_dict[id],_dict)        

c++程式碼：

/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
#include<map>
using namespace std;
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        map<int,Employee*> _map;
        for(Employee* employee:employees)
            _map[employee->id]=employee; 
        return dfs(_map[id],_map);
    }
    int dfs(Employee* root,map<int,Employee*> _map)
    {
        if(!root)
            return 0;
        int sum=root->importance;
        for (auto sub:root->subordinates)
            sum+=dfs(_map[sub],_map);
        return sum;
    }
};

優化後的c++程式碼，速度變快很多。

/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
#include<map>
using namespace std;
class Solution {
public:
    map<int,Employee*> _map;
    int getImportance(vector<Employee*> employees, int id) {
        
        for(Employee* employee:employees)
            _map[employee->id]=employee; 
        return dfs(id);
    }
    int dfs(int id)
    {
        Employee* root=_map[id];
        int sum=root->importance;
        for (auto sub:root->subordinates)
            sum+=dfs(sub);
        return sum;
    }
};

總結： 因為給的是個類，所以可以申請一個共有變數供所有函式使用，這樣傳引數的時候就不需要太多形式引數，可以簡化程式碼。才發現c++原來也可以這麼搞。