三星研究院机试（Optimal Path）

Mr. Kim has to deliver refrigerators to N customers. From the office, he is going to visit all the customers and then return to his home. Each location of the office, his home, and the customers is given in the form of integer coordinates (x,y) (0≤x≤100, 0≤y≤100) . The distance between two arbitrary locations (x1, y1) and (x2, y2) is computed by |x1-x2| + |y1-y2|, where |x| denotes the absolute value of x; for instance, |3|=|-3|=3. The locations of the office, his home, and the customers are all distinct. You should plan an optimal way to visit all the N customers and return to his among all the possibilities.

You are given the locations of the office, Mr. Kim’s home, and the customers; the number of the customers is in the range of 5 to 10. Write a program that, starting at the office, finds a (the) shortest path visiting all the customers and returning to his home. Your program only have to report the distance of a (the) shortest path.

You don’t have to solve this problem efficiently. You could find an answer by looking up all the possible ways. If you can look up all the possibilities well, you will get a perfect score.

[Constraints]

5≤N≤10. Each location (x,y) is in a bounded grid, 0≤x≤100, 0≤y≤100, and x, y are integers.

[Input]

You are given 10 test cases. Each test case consists of two lines; the first line has N, the number of the customers, and the following line enumerates the locations of the office, Mr. Kim’s home, and the customers in sequence. Each location consists of the coordinates (x,y), which is reprensented by ‘x y’.

[Output]

Output the 10 answers in 10 lines. Each line outputs the distance of a (the) shortest path. Each line looks like ‘#x answer’ where x is the index of a test case. ‘#x’ and ‘answer’ are separated by a space.

[I/O Example]

Input (20 lines in total. In the first test case, the locations of the office and the home are (0, 0) and (100, 100) respectively, and the locations of the customers are (70, 40), (30, 10), (10, 5), (90, 70), (50, 20).)

思路：典型的DFS问题，重点关注客户位置，访问路径全排列，再把最短的距离找出来。

public class Solution{

    /**
     * 5 ← Starting test case #1
     *
     * 0 0 100 100 70 40 30 10 10 5 90 70 50 20
     *
     * @param args
     */
    public static int count;
    public static int ans;
    public static int des;


    public static boolean[] visited; // 访问标记
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int T;
        T = sc.nextInt();
        for (int t = 1; t <= T; t++) {

            count = sc.nextInt();
            int nLocations = count + 2;

            int[][] locations = new int[nLocations][2];

            // 读取地点坐标
            for (int i = 0; i < nLocations; i++) {

                locations[i][0] = sc.nextInt();
                locations[i][1] = sc.nextInt();
            }
            ans = Integer.MAX_VALUE;
            des = 0;
            visited = new boolean[count];

            dfs(locations,0, 0);

            StringBuilder sb = new StringBuilder();
            sb.append("#").append(t).append(" ").append(ans);
            System.out.println(sb);

        }
    }

    public static int destination(int[] i, int[] j) {

        int des = Math.abs(j[1]-i[1])+Math.abs(j[0]-i[0]);

        return des;

    }

    private static void dfs(int[][] locations,int k ,int index) { // k 表示上一个地点的位置

        if (index ==count) {
            ans = Math.min(ans, des);
            return;
        }


        for (int i = 0; i < count; i++) {
            if (visited[i] == true) {
                continue;
            }

            visited[i] = true;
            // 计算距离
            if (index == 0) {
                des += destination(locations[0],locations[i + 2]);
            }else{
                des += destination(locations[k + 2], locations[i + 2]);
            }
            if (index == count - 1) {
                des += destination(locations[i+2],locations[1]);
            }

            dfs(locations, i,index+1);

            visited[i] = false;
            // 还原距离
            if (index == 0) {
                des -= destination(locations[0],locations[i + 2]);
            }else{
                des -= destination(locations[k + 2], locations[i + 2]);
            }
            if (index == count - 1) {
                des -= destination(locations[i+2],locations[1]);
            }

        }

    }

}