690. Employee Importance

问题：

给定一个公司的上下级关系

[id, importance, [subordinates]]

[本员工id，本员工权值，[本员工的下属们的id]]

求给定员工id，的所有下属员工+自己的权值。

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：BFS（广度优先搜索Breadth first search）

queue: 保存当前处理 id

visited：保存已经处理过的 id

遍历queue

对当前每一个 id：cur，权值相加 res+=cur->importance;

再将当前 id 的下属们，（没有遍历过的）加入queue中：queue.push(cur->subordinates[x])

代码参考：

/*
// Definition for Employee.
class Employee {
public:
    int id;
    int importance;
    vector<int> subordinates;
};
*/

class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        int res = 0;
        queue<int> q;
        unordered_set<int> visited;
        q.push(id);
        visited.insert(id);
        while(!q.empty()) {
            int sz = q.size();
            for(int i=0; i<sz; i++) {
                int cur = q.front();
                q.pop();
                for(Employee* e:employees) {
                    if(e->id == cur) {
                        res += e->importance;
                        for(int sub:e->subordinates) {
                            if(visited.insert(sub).second==true) {
                                q.push(sub);
                            }
                        }
                        break;
                    }
                }
            }
        }
        return res;
    }
};