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.------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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 = 0; 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; } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10712898.html
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