bidirection bfs

https://blog.csdn.net/hzt12345hf/article/details/79137318 knight巡逻

给一个infinite board，起点终点，国际象棋的king，问怎么得到最短距离。bfs，在追问下想到O(1)。然后棋子改成knight，问最短距离。我说了bfs，然后说可以双向bfs优化，他问can you do better than that?想不出，最后让我写了个简单的bfs。想问这个knight可以减枝吗？对国际象棋真的不是很了解。

import java.io.*;
import java.util.*;
import java.text.SimpleDateFormat;

class Solution {
  public static void main(String[] args) throws InterruptedException {
    Solution s = new Solution();
    int[][] array = new int[][] {{0,1},{1,3}};
    System.out.println(s.count(array) - 2);
  }
  static class Point {
    int x,y; 
    Point(int x, int y) {
      this.x = x;
      this.y = y;
    }
  }
  int[][][] board;

  public int count(int[][] points) {
    board = new int[50][50][2];
    Queue<Point> q1 = new LinkedList<>();
    Queue<Point> q2 = new LinkedList<>();
    Point p1 = new Point(points[0][0],points[0][1]);
    q1.offer(p1);
    board[points[0][0]][points[0][1]][0] = 1;
    Point p2 = new Point(points[1][0],points[1][1]);
    board[points[1][0]][points[1][1]][1] = 1;
    q2.offer(p2);
    int[] prevCount = new int[] {1,1};

    while (!q1.isEmpty() && !q2.isEmpty()) {
      if (q1.size() < q2.size()) {
        int a1 = work(q1,prevCount,0);
        if (a1 != -1) return a1;
      } else {
        int a2 = work(q2,prevCount,1);
        if (a2 != -1) return a2;
      }

    }
    return -1;
  }
  int[][] dir = new int[][] {{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};
  public int work(Queue<Point> q, int[] prevCount, int index){
    int count = 0;
    for (int i = 0; i < prevCount[index]; i++) {
        Point cur = q.poll();
        for (int j = 0; j < 8; j++) {
          int x = cur.x + dir[j][0];
          int y = cur.y + dir[j][1];

          if (x < 0 || x >= 30 || y < 0 || y >= 30) continue;
          if  (board[x][y][0] != 0 && board[x][y][1] != 0) {
              return board[x][y][1] + board[x][y][0];
          }
          if (board[x][y][index] != 0) continue;


          count++;
          board[x][y][index] = board[cur.x][cur.y][index] + 1;       

          Point p = new Point(x,y);
          q.offer(p);

        }
    }
    prevCount[index] = count;
    return -1;
  }




}