蓝桥杯 - 兰顿蚂蚁 Langton's ant


蓝桥杯 - 兰顿蚂蚁 - 永不冷�龅娜松�
兰顿蚂蚁,是于1986年,由克里斯・兰顿提出来的,属于细胞自动机的一种。
平面上的正方形格子被填上黑色或白色。在其中一格正方形内有一只"蚂蚁"。
蚂蚁的头部朝向为:上下左右其中一方。
蚂蚁的移动规则十分简单:
若蚂蚁在黑格,右转90度,将该格改为白格,并向前移一格;
若蚂蚁在白格,左转90度,将该格改为黑格,并向前移一格。
规则虽然简单,蚂蚁的行为却十分复杂。刚刚开始时留下的路线都会有接近对称,像是会重复,但不论起始状态如何,蚂蚁经过漫长的混乱活动后,会开辟出一条规则的"高速公路"。
蚂蚁的路线是很难事先预测的。
你的任务是根据初始状态,用计算机模拟兰顿蚂蚁在第n步行走后所处的位置。
【数据格式】
输入数据的第一行是 m n 两个整数(3 < m, n < 100),表示正方形格子的行数和列数。
接下来是 m 行数据。
每行数据为 n 个被空格分开的数字。0 表示白格,1 表示黑格。
接下来是一行数据:x y s k, 其中x y为整数,表示蚂蚁所在行号和列号(行号从上到下增长,列号从左到右增长,都是从0开始编号)。s 是一个大写字母,表示蚂蚁头的朝向,我们约定:上下左右分别用:UDLR表示。k 表示蚂蚁走的步数。
输出数据为两个空格分开的整数 p q, 分别表示蚂蚁在k步后,所处格子的行号和列号。
例如, 输入:
5 6
0 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
2 3 L 5
程序应该输出:
1 3
再例如, 输入:
3 3
0 0 0
1 1 1
1 1 1
1 1 U 6
程序应该输出:
0 0
public class Main {

//当前位置
private static int x;
private static int y;

private static int [][]a = new int[100][100];

//当前方向
private static int direction = 0;


public static void main(String[] args) {

Scanner cin = new Scanner(System.in);
int m,n;

m = cin.nextInt();
n = cin.nextInt();

for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
a[i][j] = cin.nextInt();
}

x = cin.nextInt();
y = cin.nextInt();

//将方向设置为数字  方向 U R D L
String dirtemp = cin.next();
if(dirtemp.equals("R"))
direction = 1;
else if(dirtemp.equals("D"))
direction = 2;
else if(dirtemp.equals("L"))
direction = 3;

//步数
int k = cin.nextInt();

for(int i=0;i<k;i++)
move();

System.out.print(x+" "+y);
}


// 移动的方法
public static void move() 
{
//如果是白的,则左转
if(a[x][y] == 0)
{
direction = (direction - 1) ;
if(direction < 0)
direction += 4;
a[x][y] = 1;
}

//如果是黑的则右转
else if(a[x][y] == 1)
{
direction = (direction + 1)%4;
a[x][y] = 0;
}

switch(direction)
{
case 0:
x--;
break;
case 1:
y++;
break;
case 2:
x++;
break;
case 3:
y--;
break;
}
}
}
https://en.wikipedia.org/wiki/Langton%27s_ant
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