## 题目描述

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:

1. One employee has at most one direct leader and may have several subordinates.
2. The maximum number of employees won't exceed 2000.

## 解法一

### Python

``````"""
# Employee info
class Employee:
def __init__(self, id, importance, subordinates):
# It's the unique id of each node.
# unique id of this employee
self.id = id
# the importance value of this employee
self.importance = importance
# the id of direct subordinates
self.subordinates = subordinates
"""

class Solution:
def getImportance(self, employees, id):
"""
:type employees: Employee
:type id: int
:rtype: int
"""

def dfs(employee, infos):
importance = employee.importance
for child in employee.subordinates:
importance += dfs(infos[child], infos)
return importance

infos = {}
for employee in employees:
infos[employee.id] = employee
return dfs(infos[id], infos)``````

### Java

``````/*
// Employee info
class Employee {
// It's the unique id of each node;
// unique id of this employee
public int id;
// the importance value of this employee
public int importance;
// the id of direct subordinates
public List<Integer> subordinates;
};
*/
class Solution {
public int getImportance(List<Employee> employees, int id) {
class Utils {
int dfs(Employee employee, Map<Integer, Employee> infos) {
int importance = employee.importance;
for (int child : employee.subordinates) {
importance += dfs(infos.get(child), infos);
}
return importance;
}
}
Map<Integer, Employee> infos = new HashMap<>();
for (Employee employee : employees) {
infos.put(employee.id, employee);
}
Utils utils = new Utils();
return utils.dfs(infos.get(id), infos);
}
}``````

### C++

``````/*
// 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) {
class Utils {
public:
int dfs(Employee *employee, unordered_map<int, Employee *> &infos) {
int importance = employee->importance;
for (int child : employee->subordinates) {
importance += dfs(infos[child], infos);
}
return importance;
}
};
unordered_map<int, Employee *> infos;
for (Employee *employee : employees) {
infos[employee->id] = employee;
}
Utils utils;
return utils.dfs(infos[id], infos);
}
};``````