C#winform打开进程所在的文件夹，如何使进程所对应的文件被选中？
我自己的方法打开选定的进程的文件位置，效果如图：


系统的任务管理器，打开同一进程文件位置，效果如下：

请问，怎样才能实现像系统的任务管理器一样的功能，打开文件位置的文件夹后自动选定对应的文件呢？
希望能给出实例代码，我会自行研究，谢谢！
[解决办法]
Explorer [/n] [/e] [(,)/root,<object>] [/select,<object>]

/n Opens a new single-pane window for the default
selection. This is usually the root of the drive Windows
is installed on. If the window is already open, a
duplicate opens.

/e Opens Windows Explorer in its default view.

/root,<object> Opens a window view of the specified object.


/select,<object> Opens a window view with the specified folder, file or
application selected.

[解决办法]
最后那句改成

Process.Start("explorer.exe","/select," + str)