java - 格雷厄姆扫描的问题

Question

目前正在将 Graham's Scan 与 Convex HUll 结合使用。我是一名学生，所以我想自己完成它，但是我一直在筛选多个网站来寻找答案。简而言之，我有我的构造函数，一个来自文件，一个随机生成，正在工作，因此我能够创建一系列点。下一步是实现快速排序，按极角排序。这是通过比较器类完成的。比较器类是我陷入困境的地方，我们被告知使用点比较和交叉比较来进行角度比较，但我很迷失。

/**
 * Use cross product and dot product to implement this method.  Do not take square roots 
 * or use trigonometric functions. See the PowerPoint notes on how to carry out cross and 
 * dot products. 
 * 
 * Call comparePolarAngle() and compareDistance(). 
 * 
 * @param p1
 * @param p2
 * @return -1 if one of the following three conditions holds: 
 *                a) p1 and referencePoint are the same point but p2 is a different point; 
 *                b) neither p1 nor p2 equals referencePoint, and the polar angle of 
 *                   p1 with respect to referencePoint is less than that of p2; 
 *                c) neither p1 nor p2 equals referencePoint, p1 and p2 have the same polar 
 *                   angle w.r.t. referencePoint, and p1 is closer to referencePoint than p2. 
 *         0  if p1 and p2 are the same point  
 *         1  if one of the following three conditions holds:
 *                a) p2 and referencePoint are the same point but p1 is a different point; 
 *                b) neither p1 nor p2 equals referencePoint, and the polar angle of
 *                   p1 with respect to referencePoint is greater than that of p2;
 *                c) neither p1 nor p2 equals referencePoint, p1 and p2 have the same polar
 *                   angle w.r.t. referencePoint, and p1 is further to referencePoint than p2. 
 *                   
 */
public int compare(Point p1, Point p2){
    if(p1 == referencePoint && p2 != referencePoint){
        return -1;
    } else if(p1 == p2){
        return 0;
    } else {

    }
    return 0; 
}


/**
 * Compare the polar angles of two points p1 and p2 with respect to referencePoint.  Use 
 * cross products.  Do not use trigonometric functions. 
 * 
 * Precondition:  p1 and p2 are distinct points. 
 * 
 * @param p1
 * @param p2
 * @return   -1  if p1 equals referencePoint or its polar angle with respect to referencePoint
 *               is less than that of p2. 
 *            0  if p1 and p2 have the same polar angle. 
 *            1  if p2 equals referencePoint or its polar angle with respect to referencePoint
 *               is less than that of p1. 
 */
public int comparePolarAngle(Point p1, Point p2){
    // TODO 
    return 0; 
}


/**
 * Compare the distances of two points p1 and p2 to referencePoint.  Use dot products. 
 * Do not take square roots. 
 * 
 * @param p1
 * @param p2
 * @return   -1   if p1 is closer to referencePoint 
 *            0   if p1 and p2 are equidistant to referencePoint
 *            1   if p2 is closer to referencePoint
 */
public int compareDistance(Point p1, Point p2){
    int distance = 0;

    return distance; 
}

这就是全部，我只是在陷入困境之前在比较方法上经历了一些小事情。

快速排序和分区方法非常标准，但我将添加它们，以便你们可以广泛了解所有内容:

/**
 * Sort the array points[] in the increasing order of polar angle with respect to lowestPoint.  
 * Use quickSort.  Construct an object of the pointComparator class with lowestPoint as the 
 * argument for point comparison.  
 * 
 * Ought to be private, but is made public for testing convenience.   
 */
public void quickSort(){

    // TODO 
}


/**
 * Operates on the subarray of points[] with indices between first and last. 
 * 
 * @param first  starting index of the subarray
 * @param last   ending index of the subarray
 */
private void quickSortRec(int first, int last){

    // TODO
}


/**
 * Operates on the subarray of points[] with indices between first and last.
 * 
 * @param first
 * @param last
 * @return
 */
private int partition(int first, int last){

    // TODO 
    return 0; 
}

我知道我本质上需要先启动并运行 Compare 类，然后才能开始快速排序方法，但我觉得我根本不知道如何使用点/交叉比较，所以我感觉真的迷失了.

如果有人愿意帮忙，我将不胜感激!非常感谢您的浏览，祝您度过一个美好的夜晚。

Answer 1

在所有这些方法中，当您需要查看两个 Point 对象是否相等时，您应该使用 Point 的 equals 方法，而不是“==”:

if(p1.equals(p2)) {
    //code
}

实现比较

请注意，您的比较方法需要在其实现中使用 equals()、comparePolarAngle() 和 CompareDistance()。最后一组条件(返回 1)也可以在 else 语句中处理。

public int compare(Point p1, Point p2) {
   if(p1.equals(p2)) {
      return 0;
   }
   else if(p1.equals(referencePoint) ||
          (!p1.equals(referencePoint) && !p2.equals(referencePoint) && comparePolarAngle(p1, p2) == -1) ||
          (!p1.equals(referencePoint) && !p2.equals(referencePoint) && comparePolarAngle(p1, p2) == 0 && compareDistance(p1, p2) == -1))
   {
      return -1;
   }
   else {
      return 1;
   }
}

实现比较距离

这里我们需要的主要信息是如何仅使用点积来确定从referencePoint到Point对象的 vector 的长度。首先，让我们实现一个辅助方法，该方法将两个点作为输入并以整数值形式返回点积。

private int dotProduct(Point p1, Point p2) {
   int p1X = p1.getX() - referencePoint.getX();
   int p1Y = p1.getY() - referencePoint.getY();
   int p2X = p2.getX() - referencePoint.getX();
   int p2Y = p2.getY() - referencePoint.getY();
   //compensate for a reference point other than (0, 0)

   return (p1X * p2X) + (p1Y * p2Y);  //formula for dot product
}

那么我们如何使用它来计算 vector 的长度呢？如果我们将一个点与其自身进行点积，我们会得到 (xx) + (yy)，这是毕达哥拉斯定理 (a^2 + b^2 = c^) 的左侧2)。因此，如果我们调用 dotProduct(p1, p1)，我们将得到其 vector 长度的平方。现在让我们实现compareDistance。

public int compareDistance(Point p1, Point p2) {
   if(dotProduct(p1, p1) == dotProduct(p2, p2)) {
      return 0;
   }
   else if(dotProduct(p1, p1) < dotProduct(p2, p2)) {
      return -1;
   }
   else {
      return 1;
   }
}

不需要对点积求平方根，只需比较长度的平方即可。另请注意，此处可以使用“==”，因为我们比较的是整数，而不是点。

实现comparePolarAngle

与点积一样，让我们​​实现一个计算两个输入点的叉积的辅助方法。

private int crossProduct(Point p1, Point p2) {
   int p1X = p1.getX() - referencePoint.getX();
   int p1Y = p1.getY() - referencePoint.getY();
   int p2X = p2.getX() - referencePoint.getX();
   int p2Y = p2.getY() - referencePoint.getY();
   //compensate for a reference point other than (0, 0)

   return (p1X * p2Y) - (p2X * p1Y);  //formula for cross product
}

另一种写出两点叉积结果的方法是 |p1||p2|sin(theta)，其中 |p1|是 p1 vector 的长度，|p2|是 p2 vector 的长度，theta 是 p1 到 p2 的角度。

相对于引用点具有相同极角的两个点的 theta 值为零。 sin(0) = 0，所以极角相同的两点的叉积为零。

如果p1相对于引用点的极角小于p2的极角，则p1到p2的角度为正。对于 0 < theta < 180，sin(theta) 为正。因此，如果p1和p2的叉积为正，则p1的极角一定小于p2的极角。

如果p1相对于引用点的极角大于p2的极角，则p1到p2的角度将为负。对于 -180 < theta < 0，sin(theta) 为负数。因此，如果p1和p2的叉积为负，则p1的极角一定大于p2的极角。

利用这些信息，我们最终可以实现comparePolarAngle。

public int comparePolarAngle(Point p1, Point p2) {
   if(crossProduct(p1, p2) == 0) {
      return 0;
   }
   else if(p1.equals(referencePoint) || crossProduct(p1, p2) > 0) {
      return -1;
   }
   else {
      return 1;
   }
}

我将把快速排序的实现留给你，因为我不知道你的 Point 对象是如何存储、访问和比较的。