Chapter 8 Exercise 21, Introduction to Java Programming, Tenth Edition Y. Daniel LiangY.

*8.21 (Central city) Given a set of cities, the central city is the city that has the shortest total distance to all other cities. Write a program that prompts the number of the cities and the locations of the cities (coordinates), the central city and its total distance to all other cities.

Enter the number of cities: 5
Enter the coordinates of the cities:
2.5 5 5.1 3 1 9 5.4 54 5.5 2.1
The central city is at (2.5, 5.0)
The total distance to all other cities is 60.81

import java.util.Scanner;
 
import jdk.nashorn.internal.runtime.arrays.NumericElements;
 
 
 
public class ProgrammingEx8_21 {
 
 public static void main(String[] args) {
  Scanner input = new Scanner(System.in);
  System.out.println("Enter the number of cities: ");
  int numberOfCities = input.nextInt();
  double[][] n = new double[numberOfCities][2];
  System.out.println("Enter the coordinates of the cities:");
 
  for (int i = 0; i < n.length; i++) {
   n[i][0] = input.nextDouble();
   n[i][1] = input.nextDouble();
  }
  double[] sumOfDis = calSomeOfDis(calDis(n));
  int cityIndex = findMin(sumOfDis);
   
  System.out.println("The central city is at (" + n[cityIndex][0] + "," +  n[cityIndex][1] + ")");
  System.out.println("The total distance to all other cities is " + sumOfDis[cityIndex]);
 }
 
 public static double[][] calDis(double[][] n) {
 
  // Calculate distances from cities and put them in square matrix.
  double[][] distances = new double[n.length][n.length];
  for (int i = 0; i < distances.length; i++) {
   for (int j = 0; j < distances.length; j++) {
    double x1 = n[i][0];
    double x2 = n[j][0];
    double y1 = n[i][1];
    double y2 = n[j][1];
    distances[i][j] = Math.sqrt(Math.pow(x1 - x2, 2)
      + Math.pow(y1 - y2, 2));
   }
  }
  return distances;
 }
 
 public static double[] calSomeOfDis(double[][] n) {
 
  double[] sum = new double[n.length];
 
  for (int i = 0; i < sum.length; i++) {
   for (int j = 0; j < sum.length; j++) {
    sum[i] += n[i][j];
   }
 
  }
 
  return sum;
 
 }
 
 public static int findMin(double[] n) {
 
  double min = n[0];
  int minIndex = 0;
  for (int i = 0; i < n.length; i++) {
   if (min > n[i]) {
    min = n[i];
    minIndex = i;
   }
  }
  return minIndex;
 
 }
 
}