C# 返回层次结构中的Boss--尝试应用深度优先搜索
XYZ是一家拥有首席执行官比尔和一系列员工的公司。员工可以有一个向他们报告的其他员工的列表，这些员工自己也可以有报告，等等。至少有一份报告的员工称为经理。 请使用ClosesCommonManager方法查找离两名员工最近的经理（即离CEO最远的经理）。您可以假设所有员工最终都向首席执行官报告 样本数据： 首席执行官比尔有3名员工向他汇报：{Dom，Samir，Michael} 多姆有三份报告{彼得、鲍勃、波特} 萨米尔没有报告{}迈克尔没有报告{} 彼得有两份报告{米尔顿，尼娜}鲍勃没有报告{} 波特没有报告米尔顿没有报告 尼娜没有报告{} 电话样本： 密室经理（米尔顿，尼娜）=彼得 ClosesCommonManager（尼娜，波特）=Dom closestCommonManager（尼娜，萨米尔）=比尔 密室经理（彼得，尼娜）=彼得 现在，为了解决这个问题，我已经这样做了——但我还没有找到解决办法。 我曾尝试使用简单的DFS算法，但无法完成解决方案C# 返回层次结构中的Boss--尝试应用深度优先搜索,c#,algorithm,data-structures,graph,depth-first-search,C#,Algorithm,Data Structures,Graph,Depth First Search,XYZ是一家拥有首席执行官比尔和一系列员工的公司。员工可以有一个向他们报告的其他员工的列表，这些员工自己也可以有报告，等等。至少有一份报告的员工称为经理。 请使用ClosesCommonManager方法查找离两名员工最近的经理（即离CEO最远的经理）。您可以假设所有员工最终都向首席执行官报告 样本数据： 首席执行官比尔有3名员工向他汇报：{Dom，Samir，Michael} 多姆有三份报告{彼得、鲍勃、波特} 萨米尔没有报告{}迈克尔没有报告{} 彼得有两份报告{米尔顿，尼娜}鲍勃没有报告{
public static Employee closestCommonManager(Employee ceo, Employee firstEmployee, Employee secondEmployee)
{
var visited = new HashSet<Employee>();
bool firstFound = false, secondFound = false;
Stack<Employee> stack = new Stack<Employee>(); // DFS
stack.Push(ceo);
while (stack.Count != 0)
{
Employee current = stack.Pop();
IList<Employee> employeeList = current.getReports();
if (firstEmployee.getId() == current.getId())
{
firstFound = true;
}
else if (secondEmployee.getId() == current.getId())
{
secondFound = true;
}
if (firstFound && secondFound)
return current;
// Should i return previous one? how do i keep track of the
// node which i found first in hierarchy ???
Console.WriteLine(current.getName());
foreach (var employee in employeeList)
{
if (visited.Add(employee))
{
stack.Push(employee);
}
}
}
return null;
}
public static Employee closestCommonManager（员工ceo、员工第一名员工、员工第二名员工）
{
var=newhashset（）；
bool firstFound=false，secondFound=false；
堆栈=新堆栈（）；//DFS
stack.Push（首席执行官）；
while（stack.Count！=0）
{
员工当前=stack.Pop（）；
IList employeeList=current.getReports（）；
if（firstEmployee.getId（）==current.getId（））
{
firstFound=true；
}
else if（secondEmployee.getId（）==current.getId（））
{
secondFound=true；
}
如果（第一次发现和第二次发现）
回流；
//我应该返回上一个吗？我如何跟踪
//我在层次结构中首先找到的节点？？？
Console.WriteLine（current.getName（））；
foreach（员工列表中的var员工）
{
如果（已访问。添加（员工））
{
stack.Push（员工）；
}
}
}
返回null；
}
//all managers of this Employee
public List<Employee> Bosses = new List<Employee>();
//how many managers between employee and Ceo
public int DistanceFromCeo = 0;
var commonBosses = (from f in firstEmployee.Bosses
join s in firstEmployee.Bosses on f.getId() equals s.getId()
select f).ToList();
var maxLenght = commonBosses.Max(b => b.DistanceFromCeo);
return commonBosses.FirstOrDefault(b => b.DistanceFromCeo == maxLenght);
只有在层次结构中没有循环时，它才会工作。但是我想如果有循环，那么谁是最远的还不清楚，如果你在循环中运行，状态将无限增长。这看起来像是一个请求。聪明的算法通常会对树进行一些预处理。一种简单的方法是标记树中每个元素与根的距离。然后，要找到两个节点最接近的共同祖先，首先从较低的节点向上移动一个指针，使它们位于相同的深度，然后将两个指针一起向上移动，直到指针接触。如果不进行预处理，可以同时从两个节点向上移动，将看到的所有节点添加到一个集合中，并检查何时将节点添加到已经存在的节点集合中。在任何一种情况下，当您第一次遇到来自两侧的节点时，该节点是最低的共同祖先。DFS的简单解决方案 单独的任务
创建DFS函数以查找从根目录到辅助目录的路径。（列表或堆栈）
为两个工作进程调用它，然后比较从根到工作进程的路径
最后一场比赛是你的目标这是我的解决方案，但由于某些原因，它没有被接受：
public static Employee ret;
public static bool first = false;
public static bool second = false;
public static Employee closestCommonManager(Employee ceo, Employee firstEmployee, Employee secondEmployee)
{
// Implement me
if (ceo == null || firstEmployee == null || secondEmployee == null)
return null;
if (ceo.getId() == firstEmployee.getId())
first = true;
if (ceo.getId() == secondEmployee.getId())
second = true;
if (first && second)
return ceo;
else
{
foreach (Employee e in ceo.getReports())
{
if (ret == null)
{
var r = closestCommonManager(e, firstEmployee, secondEmployee);
if (r != null && ret == null)
ret = ceo;
}
else
return ret;
}
}
return null;
}
public static Employee closestCommonManager(Employee ceo, Employee firstEmployee, Employee secondEmployee)
{
// Implement me
var Manager = ceo.getReports();
Employee tempEmployee = ceo;
foreach (var HisReporter in Manager)
{
if (HisReporter.Equals(firstEmployee) || HisReporter.Equals(secondEmployee))
{
//if one of the employees are reporting to him return manager
tempEmployee = ceo;
}
else if(HisReporter.getReports().Count > 0)
{
//if non of the employees reporting to him check if his reporters having these employees
tempEmployee = closestCommonManager(HisReporter, firstEmployee, secondEmployee);
}
}
return tempEmployee;
}
}以下是Krzysztof Bardzinski建议的DFS实现
static void Main(string[] args)
{
Dictionary<int, Employee> employeeMap = new Dictionary<int, Employee>();
Employee ceo = null, firstEmployee = null, secondEmployee = null;
/*//Sample input
List<string> input = new List<string>();
input.Add("employee 1 A");
input.Add("employee 2 B");
input.Add("employee 3 C");
input.Add("employee 4 D");
input.Add("employee 5 E");
input.Add("employee 6 F");
input.Add("employee 7 G");
input.Add("employee 8 H");
input.Add("employee 9 I");
input.Add("employee 10 J");
input.Add("employee 11 K");
input.Add("employee 12 L");
input.Add("employee 13 M");
input.Add("report 1 2");
input.Add("report 1 3");
input.Add("report 2 4");
input.Add("report 2 5");
input.Add("report 2 6");
input.Add("report 5 7");
input.Add("report 5 8");
input.Add("report 6 9");
input.Add("report 6 10");
input.Add("report 3 11");
input.Add("report 3 12");
input.Add("report 3 13");
input.Add("params 1 9 13");*/
string line;
while ((line = Console.ReadLine()) != null)
{
string[] tokens = line.Split();
string type = tokens[0];
if (type.Equals("employee"))
{
int id = int.Parse(tokens[1]);
String name = tokens[2];
employeeMap.Add(id, new Employee(id, name));
}
else if (type.Equals("report"))
{
Employee manager = employeeMap[int.Parse(tokens[1])];
Employee report = employeeMap[int.Parse(tokens[2])];
manager.addReport(report);
}
else if (type.Equals("params"))
{
ceo = employeeMap[int.Parse(tokens[1])];
firstEmployee = employeeMap[int.Parse(tokens[2])];
secondEmployee = employeeMap[int.Parse(tokens[3])];
}
else
{
// ignore comments and whitespace
}
}
Employee result = closestCommonManager(ceo, firstEmployee, secondEmployee);
}
public static Employee closestCommonManager(Employee ceo, Employee firstEmployee, Employee secondEmployee)
{
Stack<Employee> firstManagers = new Stack<Employee>();
Stack<Employee> secondManagers = new Stack<Employee>();
if (ceo.getReports().Count > 0)
{
searchManager(ceo, firstEmployee, firstManagers);
searchManager(ceo, secondEmployee, secondManagers);
if (firstManagers.Contains(secondEmployee))
{
return secondEmployee; //closest manager as this is managing the first one
}
if (secondManagers.Contains(firstEmployee))
{
return firstEmployee; //closest manager as this is managing the second one
}
//check for closest match
//make them equal in length.
while (firstManagers.Count > secondManagers.Count)
{
firstManagers.Pop();
}
while (secondManagers.Count > firstManagers.Count)
{
secondManagers.Pop();
}
int checkCount = firstManagers.Count;
for (int i = 0; i < checkCount; i++)
{
if (firstManagers.Peek().getId() == secondManagers.Peek().getId())
{
return firstManagers.Pop();
}
else
{
firstManagers.Pop();
secondManagers.Pop();
}
}
}
return null;
}
private static bool searchManager(Employee manager, Employee emp, Stack<Employee> graph)
{
bool res = false;
graph.Push(manager);
foreach (Employee e in manager.getReports())
{
if (e.getId() == emp.getId())
{
res = true;
break;
}
else
{
if (e.getReports().Count > 0)
{
if (searchManager(e, emp, graph))
{
return true;
}
else
{
graph.Pop();
}
}
}
}
return res;
}
public sealed class Employee
{
private readonly int id;
private readonly string name;
private readonly List<Employee> reports;
public Employee(int id, string name)
{
this.id = id;
this.name = name;
this.reports = new List<Employee>();
}
/// <returns>
/// an integer ID for this employee, guaranteed to be unique.
/// </returns>
public int getId()
{
return id;
}
/// <returns>
/// a String name for this employee, NOT guaranteed to be unique.
/// </returns>
public string getName()
{
return name;
}
/// <returns>
/// a List of employees which report to this employee. This list may be empty, but will
/// never be null.
/// </returns>
public IList<Employee> getReports()
{
return reports;
}
/// <summary>
/// Adds the provided employee as a report of this employee.
/// </summary>
public void addReport(Employee employee)
{
reports.Add(employee);
}
}
static void Main（字符串[]args）
{
Dictionary employeeMap=新字典（）；
员工ceo=null，第一名员工=null，第二名员工=null；
/*//样本输入
列表输入=新列表（）；
输入。添加（“员工1A”）；
输入。添加（“员工2 B”）；
输入。添加（“员工3 C”）；
输入。添加（“员工4 D”）；
输入。添加（“员工5 E”）；
输入。添加（“员工6 F”）；
输入。添加（“员工7 G”）；
输入。添加（“员工8小时”）；
输入。添加（“员工9 I”）；
输入。添加（“员工10 J”）；
输入。添加（“员工11K”）；
输入。添加（“员工12 L”）；
输入。添加（“员工13M”）；
输入。添加（“报告1-2”）；
输入。添加（“报告13”）；
输入。添加（“报告2-4”）；
输入。添加（“报告2.5”）；
输入。添加（“报告2 6”）；
输入。添加（“报告5-7”）；
输入。添加（“报告5-8”）；
输入。添加（“报告6-9”）；
输入。添加（“报告6-10”）；
输入。添加（“报告3-11”）；
输入。添加（“报告3 12”）；
输入。添加（“报告3 13”）；
输入。添加（“参数1 9 13”）*/
弦线；
而（（line=Console.ReadLine（））！=null）
{
string[]tokens=line.Split（）；
字符串类型=令牌[0]；
if（类型等于（“员工”））
{
int id=int.Parse（标记[1]）；
字符串名称=令牌[2]；
添加（id，新员工（id，姓名））；
}
else if（type.Equals（“报告”））
{
Employee manager=employeeMap[int.Parse（令牌[1]）；
Employee report=employeeMap[int.Parse（令牌[2]）；
经理。添加报告（报告）；
}
else if（type.Equals（“params”））
{
ceo=employeeMap[int.Parse（令牌[1]）；
firstEmployee=employeeMap[int.Parse（令牌[2]）；
secondEmployee=employeeMap[int.Parse（令牌[3]）；
}
其他的
{
//忽略注释和空白
}
}
static void Main(string[] args)
{
Dictionary<int, Employee> employeeMap = new Dictionary<int, Employee>();
Employee ceo = null, firstEmployee = null, secondEmployee = null;
/*//Sample input
List<string> input = new List<string>();
input.Add("employee 1 A");
input.Add("employee 2 B");
input.Add("employee 3 C");
input.Add("employee 4 D");
input.Add("employee 5 E");
input.Add("employee 6 F");
input.Add("employee 7 G");
input.Add("employee 8 H");
input.Add("employee 9 I");
input.Add("employee 10 J");
input.Add("employee 11 K");
input.Add("employee 12 L");
input.Add("employee 13 M");
input.Add("report 1 2");
input.Add("report 1 3");
input.Add("report 2 4");
input.Add("report 2 5");
input.Add("report 2 6");
input.Add("report 5 7");
input.Add("report 5 8");
input.Add("report 6 9");
input.Add("report 6 10");
input.Add("report 3 11");
input.Add("report 3 12");
input.Add("report 3 13");
input.Add("params 1 9 13");*/
string line;
while ((line = Console.ReadLine()) != null)
{
string[] tokens = line.Split();
string type = tokens[0];
if (type.Equals("employee"))
{
int id = int.Parse(tokens[1]);
String name = tokens[2];
employeeMap.Add(id, new Employee(id, name));
}
else if (type.Equals("report"))
{
Employee manager = employeeMap[int.Parse(tokens[1])];
Employee report = employeeMap[int.Parse(tokens[2])];
manager.addReport(report);
}
else if (type.Equals("params"))
{
ceo = employeeMap[int.Parse(tokens[1])];
firstEmployee = employeeMap[int.Parse(tokens[2])];
secondEmployee = employeeMap[int.Parse(tokens[3])];
}
else
{
// ignore comments and whitespace
}
}
Employee result = closestCommonManager(ceo, firstEmployee, secondEmployee);
}
public static Employee closestCommonManager(Employee ceo, Employee firstEmployee, Employee secondEmployee)
{
Stack<Employee> firstManagers = new Stack<Employee>();
Stack<Employee> secondManagers = new Stack<Employee>();
if (ceo.getReports().Count > 0)
{
searchManager(ceo, firstEmployee, firstManagers);
searchManager(ceo, secondEmployee, secondManagers);
if (firstManagers.Contains(secondEmployee))
{
return secondEmployee; //closest manager as this is managing the first one
}
if (secondManagers.Contains(firstEmployee))
{
return firstEmployee; //closest manager as this is managing the second one
}
//check for closest match
//make them equal in length.
while (firstManagers.Count > secondManagers.Count)
{
firstManagers.Pop();
}
while (secondManagers.Count > firstManagers.Count)
{
secondManagers.Pop();
}
int checkCount = firstManagers.Count;
for (int i = 0; i < checkCount; i++)
{
if (firstManagers.Peek().getId() == secondManagers.Peek().getId())
{
return firstManagers.Pop();
}
else
{
firstManagers.Pop();
secondManagers.Pop();
}
}
}
return null;
}
private static bool searchManager(Employee manager, Employee emp, Stack<Employee> graph)
{
bool res = false;
graph.Push(manager);
foreach (Employee e in manager.getReports())
{
if (e.getId() == emp.getId())
{
res = true;
break;
}
else
{
if (e.getReports().Count > 0)
{
if (searchManager(e, emp, graph))
{
return true;
}
else
{
graph.Pop();
}
}
}
}
return res;
}
public sealed class Employee
{
private readonly int id;
private readonly string name;
private readonly List<Employee> reports;
public Employee(int id, string name)
{
this.id = id;
this.name = name;
this.reports = new List<Employee>();
}
/// <returns>
/// an integer ID for this employee, guaranteed to be unique.
/// </returns>
public int getId()
{
return id;
}
/// <returns>
/// a String name for this employee, NOT guaranteed to be unique.
/// </returns>
public string getName()
{
return name;
}
/// <returns>
/// a List of employees which report to this employee. This list may be empty, but will
/// never be null.
/// </returns>
public IList<Employee> getReports()
{
return reports;
}
/// <summary>
/// Adds the provided employee as a report of this employee.
/// </summary>
public void addReport(Employee employee)
{
reports.Add(employee);
}
}