• 常用十大算法(十)— 踏棋盘算法


    常用十大算法(十)— 踏棋盘算法

    博客说明

    文章所涉及的资料来自互联网整理和个人总结,意在于个人学习和经验汇总,如有什么地方侵权,请联系本人删除,谢谢!

    介绍

    • 马踏棋盘算法也被称为骑士周游问题
    • 将马随机放在国际象棋的8×8棋盘Board0~7]的某个方格中,马按走棋规则(马走日字)进行移动。要求每个方格只进入一次,走遍棋盘上全部64个方格

    思路

    • 马踏棋盘问题(骑士周游问题)实际上是图的深度优先搜索(DFS)的应用。
    • 如果使用回溯(就是深度优先搜索)来解决,假如马儿踏了53个点,如图:走到了第53个,坐标(1,0),发现已经走到尽头,没办法,那就只能回退了,查看其他的路径,就在棋盘上不停的回溯…… ,

    代码实现

    package com.atguigu.horse;
    
    import java.awt.Point;
    import java.util.ArrayList;
    import java.util.Comparator;
    
    public class HorseChessboard {
    
    	private static int X; // 列
    	private static int Y; // 行
    	
    	private static boolean visited[];
    	private static boolean finished; 
    	
    	public static void main(String[] args) {
    		X = 8;
    		Y = 8;
    		int row = 1; 
    		int column = 1; 
    		int[][] chessboard = new int[X][Y];
    		visited = new boolean[X * Y];
    		long start = System.currentTimeMillis();
    		traversalChessboard(chessboard, row - 1, column - 1, 1);
    		long end = System.currentTimeMillis();
    		System.out.println("时间: " + (end - start));
    		for(int[] rows : chessboard) {
    			for(int step: rows) {
    				System.out.print(step + "	");
    			}
    			System.out.println();
    		}
    	}
    	
    	public static void traversalChessboard(int[][] chessboard, int row, int column, int step) {
    		chessboard[row][column] = step;
    		visited[row * X + column] = true; 
    		ArrayList<Point> ps = next(new Point(column, row));
    		sort(ps);
    		while(!ps.isEmpty()) {
    			Point p = ps.remove(0);
    			if(!visited[p.y * X + p.x]) {
    				traversalChessboard(chessboard, p.y, p.x, step + 1);
    			}
    		}
    		if(step < X * Y && !finished ) {
    			chessboard[row][column] = 0;
    			visited[row * X + column] = false;
    		} else {
    			finished = true;
    		}
    	}
    	
    
    	public static ArrayList<Point> next(Point curPoint) {
    		ArrayList<Point> ps = new ArrayList<Point>();
    		Point p1 = new Point();
    		if((p1.x = curPoint.x - 2) >= 0 && (p1.y = curPoint.y -1) >= 0) {
    			ps.add(new Point(p1));
    		}
    		if((p1.x = curPoint.x - 1) >=0 && (p1.y=curPoint.y-2)>=0) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x + 1) < X && (p1.y = curPoint.y - 2) >= 0) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x + 2) < X && (p1.y = curPoint.y - 1) >= 0) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x + 2) < X && (p1.y = curPoint.y + 1) < Y) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x + 1) < X && (p1.y = curPoint.y + 2) < Y) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x - 1) >= 0 && (p1.y = curPoint.y + 2) < Y) {
    			ps.add(new Point(p1));
    		}
    		if ((p1.x = curPoint.x - 2) >= 0 && (p1.y = curPoint.y + 1) < Y) {
    			ps.add(new Point(p1));
    		}
    		return ps;
    	}
    
    	//排序
    	public static void sort(ArrayList<Point> ps) {
    		ps.sort(new Comparator<Point>() {
    			@Override
    			public int compare(Point o1, Point o2) {
    				int count1 = next(o1).size();
    				int count2 = next(o2).size();
    				if(count1 < count2) {
    					return -1;
    				} else if (count1 == count2) {
    					return 0;
    				} else {
    					return 1;
    				}
    			}
    		});
    	}
    }
    
    

    感谢

    尚硅谷

    以及勤劳的自己,个人博客GitHub

    微信公众号

  • 相关阅读:
    软件设计中的立足点
    Clojure基础
    团队凝聚力
    执行力与领导力
    工作与生活
    分离焦虑OR责任焦虑
    保持激情
    立足点
    论研发管理--开篇
    初级码农常犯错误
  • 原文地址:https://www.cnblogs.com/guizimo/p/13638668.html
Copyright © 2020-2023  润新知