格雷厄姆扫描问题
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【中文标题】格雷厄姆扫描问题【英文标题】：Trouble with Graham's Scan 【发布时间】：2015-04-05 01:42:44 【问题描述】：目前正在与 Convex HUll 一起使用 Graham's Scan。我是一名学生，所以我试图自己完成它，但是我一直在筛选多个站点以找到答案。简而言之，我有我的构造函数，一个来自文件，一个随机生成，可以工作，所以我能够创建一个点数组。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。比较器类是我卡住的地方，我们被告知使用点比较和交叉比较来进行角度比较，但我很迷茫。
/**
* 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;
就是这样，我只是在比较方法上经历了一些小事情，然后才卡住。
quickSort 和 partition 方法是相当标准的，但我会添加它们，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;
我知道我基本上需要启动并运行 Compare 类，然后才能启动快速排序方法，但我觉得我什至不知道如何使用点/交叉比较，所以我真的很迷茫.
如果有人愿意提供帮助，我将非常感激！ 非常感谢您的观看，祝您有个愉快的夜晚。
【问题讨论】：
【参考方案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 compare 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 对象的向量长度。首先，让我们实现一个辅助方法，它以两个 Points 作为输入并将点积作为整数值返回。
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向量的长度呢？如果我们将 Point 与自身进行点积，我们得到 (xx) + (yy)，这是勾股定理 (a^2 + b^2 = c^) 的左侧2）。因此，如果我们调用 dotProduct(p1, p1)，我们将得到其向量长度的平方。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;
不需要取点积的平方根，您只需比较平方长度即可。另请注意，这里可以使用“==”，因为我们比较的是整数，而不是点数。
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式是|p1||p2|sin(theta) where |p1|是 p1 向量的长度，|p2|是 p2 的向量的长度，theta 是从 p1 到 p2 的角度。
相对于参考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 相对于参考点的极角小于 p2 的极角，则 p1 到 p2 的角度为正。对于 0
如果 p1 相对于参考点的极角大于 p2 的极角，则 p1 到 p2 的角度为负。对于 -180
使用这些信息，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。
【讨论】：
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