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; }