(BFS) leetcode 690. Employee Importance

690. Employee Importance

Easy

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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:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won't exceed 2000.

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这个题是可以用BFS，也可以用DFS，不过，关于C++的类与指针的用法一度难倒了我。。。。

C++代码：

/*
// 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) {
        int res = ;
        queue<int> q{{id}};
        unordered_map<int,Employee*> m;
        for(auto e:employees) m[e->id] = e;
        while(!q.empty()){
            auto t = q.front();
            q.pop();
            res += m[t]->importance;
            for(int num:m[t]->subordinates){
                q.push(num);
            }
        }
        return res;
    }
};