标签：data get solution sea name 节点 ota result cee
#week4#
#from leetcode#
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:
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
#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
【Breadth-first Search】690. Employee Importance（easy）
标签：data get solution sea name 节点 ota result cee
原文地址：http://www.cnblogs.com/iamxiaoyubei/p/7612116.html
