【程序员必会十大算法】之骑士周游问题

骑士周游问题又叫马踏棋盘问题

public class Main {

    //定义棋盘的行数和列数
    static int X = 8;
    static int Y = 8;
    //定义棋盘上的某个点是否被访问过
    static boolean[] isVisited;
    //记录是否周游结束
    static boolean isFinished = false;

    public static void main(String[] args) {
        isVisited = new boolean[X * Y];
        int[][] chessBoard = new int[X][Y];
        //从第一行第一列开始走，第一次走算第一步，即step=1
        travelChessboard(chessBoard,0,0,1);

        //展示下chessBoard
        for (int[] row: chessBoard){
            for (int step: row){
                System.out.print(step + "     ");
            }
            System.out.println();
        }
    }

    
    public static void travelChessboard(int[][] chessBoard,int row,int column,int step){
        //假定这一点是可以走的，所以设置已访问，步数也得加上
        chessBoard[row][column] = step;//假定可以走，所以先走过去看看，设置走过去的步数
        isVisited[row * X + column] = true;//假定可以走，所以先走过去看看，设置成已访问

        //得到下一步可以走的点的集合
        ArrayList nextPoints = getNext(new Point(column, row));
        while (!nextPoints.isEmpty()){
            //首先取出一个来
            Point nextPoint = nextPoints.remove(0);
            //如果这个点没有被访问过
            if (!isVisited[nextPoint.y * X + nextPoint.x]){
                travelChessboard(chessBoard,nextPoint.y,nextPoint.x,step + 1);//这里我一开始写了step++  其实应该是step+1
            }
        }

        //如果假定失败，其实这个点是不可以走的，那么我们就进行回溯！！！
        if (step < X * Y && !isFinished){
            chessBoard[row][column] = 0;
            isVisited[row * X + column] = false;
        }else {
            isFinished = true;
        }
    }
    
    public static ArrayList getNext(Point curPoint){
        //创建结果集
        ArrayList nextPoints = new ArrayList<>();
        Point nextPoint = new Point();

        //可以走0
        if ((nextPoint.x = curPoint.x + 2) < X && (nextPoint.y = curPoint.y - 1) >= 0){
            nextPoints.add(new Point(nextPoint));
        }
        //表示马可以走1
        if ((nextPoint.x = curPoint.x + 2) < X && (nextPoint.y = curPoint.y + 1) < Y){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走2
        if ((nextPoint.x = curPoint.x + 1) < X && (nextPoint.y = curPoint.y + 2) < Y){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走3
        if ((nextPoint.x = curPoint.x - 1) >= 0 && (nextPoint.y = curPoint.y + 2) < Y){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走4
        if ((nextPoint.x = curPoint.x - 2) >= 0 && (nextPoint.y = curPoint.y + 1) < Y){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走5
        if ((nextPoint.x = curPoint.x - 2) >= 0 && (nextPoint.y = curPoint.y - 1) >= 0){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走6
        if ((nextPoint.x = curPoint.x - 1) >= 0 && (nextPoint.y = curPoint.y - 2) >= 0){
            nextPoints.add(new Point(nextPoint));
        }
        //可以走7
        if ((nextPoint.x = curPoint.x + 1) < X && (nextPoint.y = curPoint.y - 2) >= 0){
            nextPoints.add(new Point(nextPoint));
        }

        return nextPoints;
    }
}

结果略去，等结果时间太长了

主要是添加了这个方法
添加了这个方法后，可以减少回溯的次数，极大的提高效率

public static void getNextNext(ArrayList arrayList){
      //重写集合的sort方法，将其按下一步可选点数目由小到大的顺序排列，再准确一点就是非递减排序（因为有相同点）
      arrayList.sort(new Comparator() {
          @Override
          public int compare(Point o1, Point o2) {
              //得到o1的下一步可选点的数目
              int count1 = getNext(o1).size();
              //得到o2的下一步可选点的数目
              int count2 = getNext(o2).size();
              if (count1 > count2){
                  return -1;
              }else if (count1 == count2){
                  return 0;
              }else {
                  return 1;
              }
          }
      });
  }

结果

1     16     43     32     3     18     45     22     
42     31     2     17     44     21     4     19     
15     56     53     60     33     64     23     46     
30     41     58     63     54     61     20     5     
57     14     55     52     59     34     47     24     
40     29     38     35     62     51     6     9     
13     36     27     50     11     8     25     48     
28     39     12     37     26     49     10     7