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.
解法1：bfs

/* // 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) 
    {
       unordered_map<int, int> value;
        unordered_map<int, vector<int>> sub;
        for (Employee* e : employees)
        {
            value[e->id] = e->importance;
            sub[e->id] = e->subordinates;
        }

        int sum = 0;
        queue<int> next;
        next.push(id);
        while (!next.empty())
        {
            sum += value[next.front()];
            vector<int> temp = sub[next.front()];
            next.pop();
            for (int n : temp)
                next.push(n);
        }
        return sum;



    }
};

解法2：dfs

 int getImportance(vector<Employee*> employees, int id)
    {
        unordered_map<int, Employee*> em;
        for (Employee* e : employees)
            em[e->id] = e;
        return geti1(em, id);
    }
    int geti1(unordered_map<int,Employee*> employee, int id)
    {
        Employee* e = employee[id];
        int sum = e->importance;
        vector<int> temp = e->subordinates;
        for (int n : temp)
        {
            sum +=geti1(employee, n);
        }
        return sum;
    }