用java解决骑士巡游问题

 

AutoRun.java

/*** * 解决骑士巡游问题更具创意的方法之一是由J. C. Warnsdorff在1823年提出的。其规则是：骑士总是移向具有最少出口且没有到达过的方格之一。 * @author Administrator * */public class AutoRun {//int[][] map = new int[8][8];//二维数组保存走过的地图数据， //"never pass"表示没走过，"pass"表示走过int step = 1;//走过了多少个没有重复的格子int x = 7;//象开始时的x坐标int y = 0;//象开始时的y坐标public static void main(String[] args) {AutoRun run = new AutoRun();}public AutoRun(){Node root = new Node(null ,0,0);//初始化根节点Node childNode = run(root);int[][] map = childNode.getMap(); for (int i = 0; i < 8; i++) {//打印步骤 for (int j = 0; j < 8; j++) { System.out.print( (map[i][j] + " ").substring(0, 4)); } System.out.println(); } }public Node run(Node node){ if (node.getStep()==64) //如果走完了64个格子返回最后的节点，这是递归的退出条件 return node; node.canGo(node);//生成子节点 List<Node> childNodes = node.getChildNodes(); if (childNodes == null)//没子节点的话返回null return null; for (int i = 0; i < childNodes.size(); i++) {//递归调用run Node currentNode = run(childNodes.get(i)); return currentNode; } return null;}}

 

Node.java

public class Node {private int x;//节点的x坐标private int y;//节点的y坐标private int step;//走到了第几步private List<Node> childNodes = new ArrayList<Node>();// 子节点private Node fatherNode = null;//父节点 private int[][] map = new int[8][8];//走过的地图/*** * Node的构造函数，传入它的父节点和要走的，x、y的坐标 * * @param x 要走的x坐标 * @param y 要走的y坐标 * */public Node(Node fatherNode, int x, int y) {this.fatherNode = fatherNode;this.x = x;this.y = y;if (fatherNode != null) {this.step = this.fatherNode.step + 1;//增加步数for (int i = 0; i < fatherNode.map.length; i++)//把父节点的地图传给子节点for (int j = 0; j < fatherNode.map[i].length; j++)this.map[i][j] = fatherNode.map[i][j];} elsethis.step = 1;//如果是跟节点的话步数从1开始this.map[x][y] = this.step;//把node所在的地图位置的值设置为当前的步数/*测试步骤for (int i = 0; i < 8; i++) { for (int j = 0; j < 8; j++) { System.out.print( (this.map[i][j] + " ").substring(0, 4)); } System.out.println(); } System.out.println(); System.out.println(); System.out.println(); */}public Node() {}/*** * 根据输入的node节点，生成它能走的子节点，并根据子节点的子节点个数从小到大排序 * * @param node 传入的节点 */public void canGo(Node node) {int x = node.getX();int y = node.getY();if (checkOrient(x,y,1)) {Node data = new Node(node, x - 2, y + 1);node.AddChildNode(data);}if (checkOrient(x,y,2)) {Node data = new Node(node, x - 1, y + 2);node.AddChildNode(data);}if (checkOrient(x,y,3)) {Node data = new Node(node, x + 1, y + 2);node.AddChildNode(data);}if (checkOrient(x,y,4)) {Node data = new Node(node, x + 2, y + 1);node.AddChildNode(data);}if (checkOrient(x,y,5)) {Node data = new Node(node, x + 2, y - 1);node.AddChildNode(data);}if (checkOrient(x,y,6)) {Node data = new Node(node, x + 1, y - 2);node.AddChildNode(data);}if (checkOrient(x,y,7)) {Node data = new Node(node, x - 1, y - 2);node.AddChildNode(data);}if (checkOrient(x,y,8)) {Node data = new Node(node, x - 2, y - 1);node.AddChildNode(data);}for (int i = 0; i < childNodes.size(); i++) for (int j = 0; j < childNodes.size() - i - 1; j++) if (childNodes.get(j).getChildNodeCount() > childNodes.get(j + 1).getChildNodeCount()) { Collections.swap(childNodes, j, j + 1); }}/*** * 检查这个方向是否可行 * @param x x坐标 * @param y y坐标 * @param orient 方向参数，1-8，8个方向 * @return */public boolean checkOrient(int x,int y,int orient){switch (orient){case 1:return checkBound(x - 2, y + 1) && !checkHavePass(x - 2, y + 1);case 2:return checkBound(x - 1, y + 2) && !checkHavePass(x - 1, y + 2);case 3:return checkBound(x + 1, y + 2) && !checkHavePass(x + 1, y + 2);case 4:return checkBound(x + 2, y + 1) && !checkHavePass(x + 2, y + 1);case 5:return checkBound(x + 2, y - 1) && !checkHavePass(x + 2, y - 1);case 6:return checkBound(x + 1, y - 2) && !checkHavePass(x + 1, y - 2);case 7:return checkBound(x - 1, y - 2) && !checkHavePass(x - 1, y - 2);case 8:return checkBound(x - 2, y - 1) && !checkHavePass(x - 2, y - 1);}return false;}/*** * 统计该节点的子节点个数 * @return 子节点个数 */public int getChildNodeCount() { int count = 0; if (checkOrient(x,y,1)) count++; if (checkOrient(x,y,2)) count++; if (checkOrient(x,y,3)) count++; if (checkOrient(x,y,4)) count++; if (checkOrient(x,y,5)) count++; if (checkOrient(x,y,6)) count++; if (checkOrient(x,y,7)) count++; if (checkOrient(x,y,8)) count++; return count; }/*** * 检查是否之前到过这个格 * * @param x 象的x坐标 * @param y 象的y坐标 * @return */private boolean checkHavePass(int x, int y) {if (map[x][y] != 0) {return true;}return false;}/*** * 检查是否越界，即坐标是否在0-7的范围内 * * @param x 象的x坐标 * @param y 象的y坐标 * @return 象是否可行 */private boolean checkBound(int x, int y) {if (x >= 0 && x < 8 && y >= 0 && y < 8)return true;return false;}/*** * 增加子节点 * * @param stepData */public void AddChildNode(Node node) {childNodes.add(node);}public List<Node> getChildNodes() {return childNodes;}public void setChildNodes(List<Node> childNodes) {this.childNodes = childNodes;}public Node getFatherNode() {return fatherNode;}public void setFatherNode(Node fatherNode) {this.fatherNode = fatherNode;}public int getX() {return x;}public void setX(int x) {this.x = x;}public int getY() {return y;}public void setY(int y) {this.y = y;}public int getStep() {return step;}public void setStep(int step) {this.step = step;}public int[][] getMap() {return map;}public void setMap(int[][] map) {this.map = map;}}