|
|
/*/*******************************************************************************
|
|
|
** **
|
|
|
** Jiedi(China nanjing)Ltd. **
|
|
|
** 创建:丁宋涛 夏曹俊,此代码可用作为学习参考 **
|
|
|
*******************************************************************************/
|
|
|
|
|
|
/*****************************FILE INFOMATION***********************************
|
|
|
**
|
|
|
** Project : 算法设计与编程实践---基于leetcode的企业真题库
|
|
|
** Contact : xiacaojun@qq.com
|
|
|
** 博客 : http://blog.csdn.net/jiedichina
|
|
|
** 视频课程 : 网易云课堂 http://study.163.com/u/xiacaojun
|
|
|
腾讯课堂 https://jiedi.ke.qq.com/
|
|
|
csdn学院 https://edu.csdn.net/course/detail/25037
|
|
|
** 51cto学院 http://edu.51cto.com/lecturer/index/user_id-100013755.html
|
|
|
** 老夏课堂 http://www.laoxiaketang.com
|
|
|
**
|
|
|
** 算法设计与编程实践---基于leetcode的企业真题库 课程群 :296249312 加入群下载代码和交流
|
|
|
** 微信公众号 : jiedi2007
|
|
|
** 头条号 : 夏曹俊
|
|
|
**
|
|
|
*****************************************************************************
|
|
|
//!!!!!!!!! 算法设计与编程实践---基于leetcode的企业真题库 课程 QQ群:296249312 下载代码和交流*/
|
|
|
/*
|
|
|
* @lc app=leetcode.cn id=690 lang=cpp
|
|
|
*
|
|
|
* [690] 员工的重要性
|
|
|
*/
|
|
|
/*
|
|
|
// 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;
|
|
|
};
|
|
|
*/
|
|
|
#include <vector>
|
|
|
#include <queue>
|
|
|
using namespace std;
|
|
|
class Solution {
|
|
|
public:
|
|
|
int getImportance(vector<Employee*> employees, int id) {
|
|
|
int res = 0;
|
|
|
for(auto k :employees){
|
|
|
if(k->id == id){
|
|
|
BFS(employees,k,res);
|
|
|
}
|
|
|
}
|
|
|
return res;
|
|
|
}
|
|
|
void BFS(vector<Employee*> employees,Employee* k,int& res){
|
|
|
queue<Employee*> q;
|
|
|
q.push(k);
|
|
|
while(!q.empty()){
|
|
|
Employee* top = q.front();
|
|
|
q.pop();
|
|
|
res += top->importance;
|
|
|
if(top->subordinates.size()!=0){
|
|
|
for(auto k:top->subordinates){
|
|
|
for(int i=0;i<employees.size();i++){
|
|
|
if(k==employees[i]->id){
|
|
|
q.push(employees[i]);
|
|
|
break;
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
}
|
|
|
};
|
|
|
|