Breadth-first Search690. Employee Importance（easy）

Posted 2020-10-09 小预备

#week4#

#from leetcode#

Description

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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 directsubordinates‘ 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.

 

Analysis

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 每一个employee看做一个节点，有权值importance——点集

employee到自己的直接下属有一条边edge——边集

然后利用BFS从id的employee开始遍历自己的下属，同时用total保存importance的和

最终的total值便是结果

 

Code

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#include <vector>
#include <queue>
using namespace std;

class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        int size = employees.size();
        int* employees_value = new int[size+1];
        vector<pair<int, int>> edge;
        queue<int> que;
        bool* hasDone = new bool[size+1];
        int total;
        // 遍历所有的employees
        // 得到边的集合以及点的value
        for (vector<Employee*>::iterator i = employees.begin();
            i != employees.end(); i++) {
            Employee* employee = (*i);
            cout << employee->id;
            hasDone[employee->id] = false;
            employees_value[employee->id] = employee->importance;
            if (!(employee->subordinates).empty())
                for (int j = 0; j < (employee->subordinates).size(); j++)
                    edge.push_back(make_pair(employee->id, employee->subordinates[j]));
        }

        // bfs
        que.push(id);
        hasDone[id] = true;
        total = employees_value[id];
        while (!que.empty()) {
            int top = que.front();
            que.pop();
            for (vector<pair<int, int>>::iterator i = edge.begin(); i != edge.end(); i++) {
                if ((*i).first == top && !(hasDone[(*i).second])) {
                    int j = ((*i).second);
                    que.push(j);
                    hasDone[j] = true;
                    total += employees_value[j];
                }
            }
        }
        return total;
    }
};

 

Runtime: 15 ms