11. 跳马问题

有一个无限大的棋盘,棋盘上有一匹马,马移动的方式为日字型。即假设马当前的位置为(x,y),那么下一步可以移动到(x+1,y+2),(x+1,y-2),(x-1,y+2),(x-1,y-2),(x+2,y+1),(x+2,y-1),(x-2,y+1)或者(x-2,y-1)这8个位置。

问马是否能从坐标(x,y)按照上述移动规则移动到坐标(x2,y2)。

 

11. 跳马问题
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication9
{
    class Program
    {
        static void Main(string[] args)
        {
            Coordinate a = new Coordinate(4, 4);
            Coordinate b = new Coordinate(6, 6);
            Console.Write(IsAbleMoveTo(a, b));
        }

        static Coordinate MoveUpOne(Coordinate a)
        {
            a.x += 2;
            a.y += 1;
            a.x -= 1;
            a.y -= 2;
            a.x -= 1;
            a.y += 2;
            return a;
        }

        static Coordinate MoveDownOne(Coordinate a)
        {
            a.x += 1;
            a.y -= 2;
            a.x += 1;
            a.y += 2;
            a.x -= 2;
            a.x -= 1;
            return a;
        }

        static Coordinate MoveLeftOne(Coordinate a)
        {
            a.x -= 1;
            a.y += 2;
            a.x += 2;
            a.y -= 1;
            a.x -= 2;
            a.y -= 1;
            return a;
        }

        static Coordinate MoveRightOne(Coordinate a)
        {
            a.x += 2;
            a.y += 1;
            a.x -= 2;
            a.y += 1;
            a.x += 1;
            a.y -= 2;
            return a;
        }

        static bool IsAbleMoveTo(Coordinate a, Coordinate b)
        {
            int x = b.x - a.x;
            int y = b.y - a.y;
            if (x>0)
            {
                for (int i = 0; i < x; i++)
                {
                    a=MoveRightOne(a);
                }
            }
            else
            {
                for (int i = 0; i < -x; i++)
                {
                    a=MoveLeftOne(a);
                }
            }
            if (y>0)
            {
                for (int i = 0; i < y; i++)
                {
                    a=MoveUpOne(a);
                }
            }
            else
            {
                for (int i = 0; i < -y; i++)
                {
                    a=MoveDownOne(a);
                }
            }
            if (a.x == b.x && a.y == b.y)
            {
                return true;
            }
            else
            {
                return false;
            }
        }
        
    }

    struct Coordinate
    {
        public int x;
        public int y;
        public Coordinate(int a, int b)
        {
            x = a;
            y = b;
        }
    }
}
View Code

11. 跳马问题

上一篇:初入职场


下一篇:POJ 2004 Mix and Build (预处理+dfs)