690. Employee Importance（BFS）

1. Description

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, []]], 1Output: 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.

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2. Naive Analysis

很简单的一道题目，找到源结点，根据广度优秀搜索遍历就行了。因为算法中涉及到每一个相关id需要查找到对应结点，所以算法的时间复杂度为 O(V2)，V为结点数。

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3. Naive Solution

/*// Employee infoclass 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:    /*    * 根据id找到employee    */    static Employee* selectById(vector<Employee*>& employees, int id) {        for(int i = 0; i < employees.size(); i++) {            if((employees[i])->id == id) {                return employees[i];            }        }        return NULL;    }     /*     * 先找到出发的结点，再根据此结点展开广度优先搜索，     * 因为每次都需要查找结点，所以时间复杂度为O(n^2)     */    int getImportance(vector<Employee*> employees, int id) {        queue<Employee*> tmp;        Employee *ptr = NULL, *source = NULL, *sub = NULL;        int count = 0;        source  =   selectById(employees, id);        if(source != NULL)            tmp.push(source);        while(!tmp.empty()) {            ptr = tmp.front();            tmp.pop();            count += ptr->importance;            vector<int> t_vec = ptr->subordinates;            for(int j = 0; j < t_vec.size(); j++) {                sub = selectById(employees, t_vec[j]);                if(sub != NULL)                    tmp.push(sub);            }        }        return count;    }};

4. 优化

如果给定的employees向量是按id排序的，那么可以直接存取，那么就无须维护一个Employee指正的队列而转而维护一个Id的队列，这样不就在空间，时间上也得到了优化。

class Solution {public:    int getImportance(vector<Employee*> employees, int id) {    queue<int> tmp;    int index;    int count = 0;    tmp.push(id);    while(!tmp.empty()) {        index = tmp.front();        tmp.pop();        count += (employees[index-1])->importance;        vector<int> t_vec = (employees[index-1])->subordinates;        for(int j = 0; j < t_vec.size(); j++) {                tmp.push(t_vec[j]);        }    }    return count;    }};