The problem. Given a set of N distinct points in the plane, draw every (maximal) line segment that connects a subset of 4 or more of the points.
Point data type. Create an immutable data type Point that represents a point in the plane by implementing the following API:
public class Point implements Comparable<Point> { public final Comparator<Point> SLOPE_ORDER; // compare points by slope to this point public Point(int x, int y) // construct the point (x, y) public void draw() // draw this point public void drawTo(Point that) // draw the line segment from this point to that point public String toString() // string representation public int compareTo(Point that) // is this point lexicographically smaller than that point? public double slopeTo(Point that) // the slope between this point and that point }
To get started, use the data type Point.java, which implements the constructor and the draw(), drawTo(), and toString() methods. Your job is to add the following components.
- The compareTo() method should compare points by their y-coordinates, breaking ties by their x-coordinates. Formally, the invoking point (x0, y0) is less than the argument point (x1, y1) if and only if either y0 < y1 or if y0 = y1 and x0 < x1.
- The slopeTo() method should return the slope between the invoking point (x0, y0) and the argument point (x1, y1), which is given by the formula (y1 − y0) / (x1 −x0). Treat the slope of a horizontal line segment as positive zero; treat the slope of a vertical line segment as positive infinity; treat the slope of a degenerate line segment (between a point and itself) as negative infinity.
- The SLOPE_ORDER comparator should compare points by the slopes they make with the invoking point (x0, y0). Formally, the point (x1, y1) is less than the point (x2,y2) if and only if the slope (y1 − y0) / (x1 − x0) is less than the slope (y2 − y0) / (x2 − x0). Treat horizontal, vertical, and degenerate line segments as in the slopeTo() method.
import java.util.Comparator; public class Point implements Comparable<Point> { public final Comparator<Point> SLOPE_ORDER = new PointCmp(); private final int x; // x coordinate private final int y; // y coordinate public Point(int x, int y) { /* DO NOT MODIFY */ this.x = x; this.y = y; } public void draw() { /* DO NOT MODIFY */ StdDraw.point(x, y); } public void drawTo(Point that) { /* DO NOT MODIFY */ StdDraw.line(this.x, this.y, that.x, that.y); } public double slopeTo(Point that) { /* YOUR CODE HERE */ if (this.compareTo(that) == 0) return Double.POSITIVE_INFINITY*-1; else if (this.x == that.x) return Double.POSITIVE_INFINITY; else if (this.y == that.y) return +0; else return (that.y - this.y) * 1.0 / (that.x - this.x); } private class PointCmp implements Comparator<Point> { @Override public int compare(Point o1, Point o2) { // TODO Auto-generated method stub if (slopeTo(o1) < slopeTo(o2) || (slopeTo(o1) == slopeTo(o2) && o1.compareTo(o2) == -1)) return -1; else if (slopeTo(o1) > slopeTo(o2) || (slopeTo(o1) == slopeTo(o2) && o1.compareTo(o2) == 1)) return 1; else return 0; } } @Override public int compareTo(Point that) { // TODO Auto-generated method stub if (this.y < that.y || (this.y == that.y && this.x < that.x)) return -1; else if (this.y == that.y && this.x == that.x) return 0; else return 1; } public String toString() { /* DO NOT MODIFY */ return "(" + x + ", " + y + ")"; } public static void main(String[] args) { // TODO Auto-generated method stub In in = new In(args[0]); int num = in.readInt(); Point points[] = new Point[num]; for (int i = 0; i < num; i++) { int x = in.readInt(); int y = in.readInt(); points[i] = new Point(x, y); } } }
Brute force. Write a program Brute.java that examines 4 points at a time and checks whether they all lie on the same line segment, printing out any such line segments to standard output and drawing them using standard drawing. To check whether the 4 points p, q, r, and s are collinear, check whether the slopes between p and q, between p and r, and between p and s are all equal.
The order of growth of the running time of your program should be N4 in the worst case and it should use space proportional to N.
public class Brute { public static void main(String[] args) { // rescale coordinates and turn on animation mode StdDraw.setXscale(0, 32768); StdDraw.setYscale(0, 32768); StdDraw.show(0); StdDraw.setPenRadius(0.01); // make the points a bit larger // read in the input String filename = args[0]; In in = new In(filename); int N = in.readInt(); Point points[] = new Point[N]; for (int i = 0; i < N; i++) { int x = in.readInt(); int y = in.readInt(); points[i] = new Point(x, y); points[i].draw(); } for (int i = 0; i < N; i++) { for (int j = 0; j < N - i - 1; j++) { if (points[j].compareTo(points[j+1]) == 1) { Point temp = points[j]; points[j] = points[j + 1]; points[j + 1] = temp; } } } for (int i = 0; i < N; i++) { for (int j = i + 1; j < N; j++) { for (int k = j + 1; k < N; k++) { for(int l = k + 1; l < N; l++) { if(points[i].slopeTo(points[j]) == points[j].slopeTo(points[k])&& points[j].slopeTo(points[k]) == points[k].slopeTo(points[l])) { points[i].drawTo(points[l]); StdOut.print(points[i].toString()+" -> "+points[j].toString() +" -> "+points[k].toString()+" -> "+points[l].toString()); StdOut.println(); } } } } } // display to screen all at once StdDraw.show(0); // reset the pen radius StdDraw.setPenRadius(); } }
A faster, sorting-based solution. Remarkably, it is possible to solve the problem much faster than the brute-force solution described above. Given a point p, the following method determines whether p participates in a set of 4 or more collinear points.
- Think of p as the origin.
- For each other point q, determine the slope it makes with p.
- Sort the points according to the slopes they makes with p.
- Check if any 3 (or more) adjacent points in the sorted order have equal slopes with respect to p. If so, these points, together with p, are collinear.
Applying this method for each of the N points in turn yields an efficient algorithm to the problem. The algorithm solves the problem because points that have equal slopes with respect to p are collinear, and sorting brings such points together. The algorithm is fast because the bottleneck operation is sorting.
Write a program Fast.java that implements this algorithm. The order of growth of the running time of your program should be N2 log N in the worst case and it should use space proportional to N.
源代码待补；