如何檢測重疊的圓形並相應地填充顏色？

Question

我創建了5個圓形，使用3個數組（x，y和半徑大小）隨機選擇x和y座標和半徑。但是，我需要圓圈根據是否與另一個圓圈重疊來動態更改顏色。所以如果5個圓圈中的一個完全不重疊，它應該被塗成黑色。重疊的圓應該是青色。如果兩個圓的中心點之間的距離小於它們的半徑之和，則認爲兩個圓重疊。

這是我迄今爲圈子類寫的。 以下代碼將成功繪製小程序窗口中的5個圓圈，並且距離已成功計算，但問題與着色有關。顏色填充似乎存在邏輯錯誤，我在這裏看不到問題。有什麼建議麼？非常感謝。

public class Circles extends Applet { 

public void paint(Graphics page) 
{ 
    Random locator = new Random(); 
    int [] xpt = new int [5]; 
    int [] ypt = new int [5]; 
    int [] rad = new int [5]; 

    setPreferredSize (new Dimension(300, 300)); 
    for (int i = 0; i < xpt.length; i++){ 

     xpt[i] = locator.nextInt(100); //need to set a number or it goes into millions, cannot set it in Random() 
     ypt[i] = locator.nextInt(100); 
     rad[i] = locator.nextInt(100); 
     System.out.println("The #" + i + " x-point: " + xpt[i] + " y-point: " + ypt[i] + " radius: " + rad[i]); //for debugging purposes 

     for (int j = 0; j < xpt.length; j++){ 
      double xpoint1 = xpt[i]+rad[i]; 
      double ypoint1 = ypt[i]+rad[i]; 
      double xpoint2 = xpt[j]+rad[j]; 
      double ypoint2 = ypt[j]+rad[j]; 
      double radius1 = rad[i]; 
      double radius2 = rad[j]; 
      double theDistance = distance(xpoint1,ypoint1,xpoint2,ypoint2); 
      System.out.println("Comparing " + i + " to " + j); //for debugging and logic checking 
      if (i==j) 
       ; 
      else if (theDistance <= (radius1+radius2)) 
      { 
       page.setColor(Color.cyan); 
       page.fillOval(xpt[i], ypt[i], rad[i], rad[i]); 
       //page.fillOval(xpt[j], ypt[j], rad[j], rad[j]); 
       System.out.println("Overlap occurred. Colored " + i + " and " + j + " cyan."); 
       System.out.println("Center points: ("+ xpoint1 +", "+ ypoint1 +") and ("+ xpoint2 + ", "+ ypoint2 + ")."); 
      } 
      else 
      { 
       page.setColor(Color.black); 
       page.fillOval(xpt[i], ypt[i], rad[i], rad[i]); 
       //page.fillOval(xpt[j], ypt[j], rad[j], rad[j]); 
       System.out.println("No overlap. Made " + i + " and " + j + " black."); 
      } 
     } 
    } 
} 

public static double distance(
     double x1, double y1, double x2, double y2) { 
    return Math.sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)); 

} 
} 

Answer 1

爲什麼你+rad[]在這裏？您不必添加半徑來比較距離。

 double xpoint1 = xpt[i]+rad[i]; 
     double ypoint1 = ypt[i]+rad[i]; 
     double xpoint2 = xpt[j]+rad[j]; 
     double ypoint2 = ypt[j]+rad[j]; 
[...] 
     double theDistance = distance(xpoint1,ypoint1,xpoint2,ypoint2); 
[...] 
     page.fillOval(xpt[i], ypt[i], rad[i], rad[i]); 

您應該使用xpt/ypt的距離。不xpoint1，並使用-的價值和2 *半徑大小......即：

 double xpoint1 = xpt[i]-rad[i]; 
     double ypoint1 = ypt[i]-rad[i]; 
     double xpoint2 = xpt[j]-rad[j]; 
     double ypoint2 = ypt[j]-rad[j]; 
[...] 
     double theDistance = distance(xpt[i],ypt[i],xpt[j],ypt[j]); 
[...] 
     page.fillOval(xpoint1 , ypoint1, 2*rad[i], 2*rad[i]); 

Answer 2

的XPOINT，ypoint等線路是不是做你的想法。

如果要查找兩個圓是否重疊，則需要查找圓的中心之間的距離是大於還是小於它們的半徑之和。

所以：

function circlesCollide(x1, y1, r1, x2, y2, r2){ 
    return (distance(x1, y1, x2, y2) <= (r1 + r2)); 
}