Knight Moves 解题报告

  这道题是找出一个骑士去访问另一个骑士的最少移动的步数，可以用bfs解决，做这道题首先要了解马的走法，马先走直线在走斜线，共有8种走法如下图:

 

 D4可以走周围标红的位置，可以用一个数组来储存.

 算发：用bfs可以解决。

 public class Main {private static int[][] position = new int[8][2];private static int[][] chessboard = new int[8][8];public Main() {initPosition();}/** * 初始化马可以走的方向 */private static void initPosition() {position[0][0] = -2;position[0][1] = -1;position[1][0] = -2;position[1][1] = 1;position[2][0] = -1;position[2][1] = -2;position[3][0] = -1;position[3][1] = 2;position[4][0] = 1;position[4][1] = -2;position[5][0] = 2;position[5][1] = -1;position[6][0] = 1;position[6][1] = 2;position[7][0] = 2;position[7][1] = 1;}private static int[] getRC(String s) {int[] rc = new int[2];rc[0] = Integer.valueOf(String.valueOf(s.charAt(1))) - 1;rc[1] = s.charAt(0) - 'a';return rc;}/** * 广度搜索算法 * * @param s1 * @param s2 */public static void bsf(String s1, String s2) {int[] rc = getRC(s1);int formR = rc[0];int formC = rc[1];// System.out.println(formR + " ," + formC);rc = getRC(s2);int toR = rc[0];int toC = rc[1];// System.out.println(toR + " ," + toC);LinkedList<Point> queue = new LinkedList<Point>();Point p = new Point();p.r = formR;p.c = formC;p.steps = 0;queue.addLast(p);chessboard[p.r][p.c] = 1;p = null;while (!queue.isEmpty()) {p = queue.getFirst();queue.removeFirst();if (p.r == toR && p.c == toC) {System.out.println("To get from " + s1 + " to " + s2+ " takes " + p.steps + " knight moves.");break;}for (int i = 0; i < 8; ++i) {Point pp = new Point();pp.r = p.r + position[i][0];pp.c = p.c + position[i][1];pp.steps = p.steps + 1;if (pp.r >= 0 && pp.c >= 0 && pp.r <= 7 && pp.c <= 7&& chessboard[pp.r][pp.c] == 0) {queue.addLast(pp);chessboard[pp.r][pp.c] = 1;}pp = null;}p = null;}}public static void main(String[] args) {Scanner sc = new Scanner(System.in);initPosition();while (sc.hasNext()) {chessboard = null;chessboard = new int[8][8];String s1 = sc.next();String s2 = sc.next();Main.bsf(s1, s2);}}}class Point {int r;int c;int steps;}