根据winform中的鼠标单击找到最短距离的对象

Question

我在使该功能正常工作时遇到问题。我必须创建一个充当出租车映射器的Winform应用程序。在加载时，出租车根据文本文件放置在相同的位置。当用户点击表格时，最近的出租车应该移动到“用户”或位置，然后停车。

除了最近的出租车并不总是去的位置，一切工作正常。更远的出租车将转到该位置。它似乎在某些时候有效，但并非全部。我不知道如果我的逻辑是在Form1_MouseDown功能

private void Form1_MouseDown(object sender, MouseEventArgs e) 
{ 
    if (pickUp) //Are we picking up a passenger? 
    { 
     //Convert mouse pointer location to local window locations 
     int mLocalX = this.PointToClient(Cursor.Position).X; 
     int mLocalY = this.PointToClient(Cursor.Position).Y; 

     //set the minimum value (for range finding) 
     int min = int.MaxValue; 
     //Temporary object to get the handle for the taxi object we want to manipulate 
     taxiCabTmp = new TaxiClass(); 

     //Iterate through each object to determine who is the closest 
     foreach (TaxiClass taxiCab in taxi) 
     { 
      if (Math.Abs(Math.Abs(taxiCab.CabLocationX - mLocalX) + Math.Abs(taxiCab.CabLocationY - mLocalY)) <= min) 
      { 
       min = Math.Abs(Math.Abs(taxiCab.CabLocationX - mLocalX) + Math.Abs(taxiCab.CabLocationY - mLocalY)); 
       //We found a minimum, grab a handle to the object's instance 
       taxiCab.GetReference(ref taxiCabTmp); 
      } 
     } 
     //Call the propogate method so it can spin off a thread to slowly change it's location for the timer to also change 
     taxiCabTmp.Propogate(mLocalX - 20, mLocalY - 20); 
     taxiCabTmp.occupied = true; //This taxi object is occupied 
     pickUp = false; //We are not picking up a passenger at the moment 
    } 
    else //We are dropping off a passenger 
    { 
     taxiCabTmp.Propogate(this.PointToClient(Cursor.Position).X, this.PointToClient(Cursor.Position).Y); 
     taxiCabTmp.occupied = false; 
     pickUp = true; //We can pick up a passenger again! 
    } 
} 

Answer 1

使用此公式计算两个坐标之间的距离。

var distance = Math.Sqrt(Math.Pow(taxiCab.CabLocationX - mLocalX, 2) + Math.Pow(taxiCab.CabLocationY - mLocalY, 2)); 
if (distance <= min) { 
    min = distance; 
    //We found a minimum, grab a handle to the object's instance 
    taxiCab.GetReference(ref taxiCabTmp); 
} 

Answer 2

你是正确的，你用于确定距离的计算并不总是正确无误。一个角度的物体会计算得比实际距离更远。

看一看这个链接了解更多信息：http://www.purplemath.com/modules/distform.htm

下面是一个例子：

int mLocalX = 1; 
int mLocalY = 1; 
int taxiCab.CabLocationX = 2; 
int taxiCab.CabLocationY = 2; 

double distance = Math.Sqrt(Math.Pow((taxiCab.CabLocationX - mLocalX), 2) + Math.Pow((taxiCab.CabLocationY - mLocalY), 2)); 

正如一个侧面说明，你应该不是你的类与类追加，即TaxiClass，它应该简单地称为出租车。