用java分治法解决邮局选址问题

邮局选址问题是指在地图上选取若干个点作为邮局的位置，使得这些邮局到所有居民的距离之和最小。这个问题可以使用分治法来解决。

具体地，我们可以将地图分成若干个区域，对于每个区域，我们选取其中心点作为邮局的位置。然后再对每个区域递归地进行分治，直到每个区域内只有一个居民点。最后将各个区域的邮局位置合并起来，就得到了最终的邮局位置。

Java代码实现如下：

public class PostOfficeLocation {
    public static void main(String[] args) {
        double[] x = {1, 2, 4, 6, 7};
        double[] y = {1, 1, 3, 5, 7};
        Point[] points = new Point[x.length];
        for (int i = 0; i < x.length; i++) {
            points[i] = new Point(x[i], y[i]);
        }
        Point[] postOffices = findPostOfficeLocations(points, 2);
        System.out.println("The locations of post offices are:");
        for (Point p : postOffices) {
            System.out.println("(" + p.x + ", " + p.y + ")");
        }
    }

    public static Point[] findPostOfficeLocations(Point[] points, int num) {
        if (points == null || points.length == 0 || num <= 0 || num > points.length) {
            return new Point[0];
        }
        return findPostOfficeLocations(points, num, 0, points.length - 1);
    }

    private static Point[] findPostOfficeLocations(Point[] points, int num, int start, int end) {
        if (num == end - start + 1) {
            Point[] result = new Point[num];
            System.arraycopy(points, start, result, 0, num);
            return result;
        }
        Point[] left = findPostOfficeLocations(points, num, start, start + num - 2);
        Point[] right = findPostOfficeLocations(points, num, start + num - 1, end);
        double minCost = Double.MAX_VALUE;
        Point[] bestCenters = null;
        for (int i = 0; i < left.length; i++) {
            for (int j = 0; j < right.length; j++) {
                Point[] centers = new Point[num];
                centers[0] = left[i];
                centers[num - 1] = right[j];
                double cost = 0;
                for (int k = start; k <= end; k++) {
                    double minDistance = Double.MAX_VALUE;
                    for (int l = 0; l < num; l++) {
                        double distance = points[k].distance(centers[l]);
                        if (distance < minDistance) {
                            minDistance = distance;
                        }
                    }
                    cost += minDistance;
                }
                for (int l = 1; l < num - 1; l++) {
                    double minDistance = Double.MAX_VALUE;
                    for (int k = start; k <= end; k++) {
                        double distance = points[k].distance(centers[l]);
                        if (distance < minDistance) {
                            minDistance = distance;
                        }
                    }
                    cost += minDistance;
                }
                if (cost < minCost) {
                    minCost = cost;
                    bestCenters = centers;
                }
            }
        }
        return bestCenters;
    }
}

class Point {
    double x;
    double y;

    public Point(double x, double y) {
        this.x = x;
        this.y = y;
    }

    public double distance(Point p) {
        return Math.sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));
    }
}

这个算法的时间复杂度为 $O(n^2 \log n)$，其中 $n$ 是居民点的数量。因为算法使用了分治法，所以空间复杂度为 $O(n \log n)$。