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老鼠走迷宫

摘要: 老鼠走迷宫

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说明
老鼠走迷宫是递归求解的基本题型,我们在二维数组中使用2表示迷宫墙壁,使用1来表示老鼠的行走路径,试用程序求出由入口至出口的路径。
解法
老鼠的走法有上、左、下、右四个方向,在每前进一格之后就选一个方向前进,无法前进时退回选择下一个可前进方向,如此在二维数组阵列中依序测试四个方向,直到走到出口为止,这是递归的基本题,请直接看程序就可以理解。
public class Mouse {
private int startI, startJ; // 入口
private int endI, endJ; // 出口
private boolean success = false;

public static void main(String[] args) {
int[][] maze = {{2, 2, 2, 2, 2, 2, 2},
{2, 0, 0, 0, 0, 0, 2},
{2, 0, 2, 0, 2, 0, 2},
{2, 0, 0, 2, 0, 2, 2},
{2, 2, 0, 2, 0, 2, 2},
{2, 0, 0, 0, 0, 0, 2},
{2, 2, 2, 2, 2, 2, 2}};

System.out.println("显示迷宫:");
for(int i = 0; i < maze.length; i++) {
for(int j = 0; j < maze[0].length; j++)
if(maze[i][j] == 2)
System.out.print("■");
else
System.out.print(" ");
System.out.println();
}

Mouse mouse = new Mouse();
mouse.setStart(1, 1);
mouse.setEnd(5, 5);

if(!mouse.go(maze)) {
System.out.println("\n没有找到出口!");
}
else {
System.out.println("\n找到出口!");
for(int i = 0; i < maze.length; i++) {
for(int j = 0; j < maze[0].length; j++) {
if(maze[i][j] == 2)
System.out.print("■");
else if(maze[i][j] == 1)
System.out.print("◇");
else
System.out.print(" ");
}
System.out.println();
}
}
}

public void setStart(int i, int j) {
this.startI = i;
this.startJ = j;
}

public void setEnd(int i, int j) {
this.endI = i;
this.endJ = j;
}

public boolean go(int[][] maze) {
return visit(maze, startI, startJ);
}

private boolean visit(int[][] maze, int i, int j) {
maze[i][j] = 1;

if(i == endI && j == endJ)
success = true;

if(!success && maze[i][j+1] == 0)
visit(maze, i, j+1);
if(!success && maze[i+1][j] == 0)
visit(maze, i+1, j);
if(!success && maze[i][j-1] == 0)
visit(maze, i, j-1);
if(!success && maze[i-1][j] == 0)
visit(maze, i-1, j);

if(!success)
maze[i][j] = 0;

return success;
}
}



由于迷宫的设计,老鼠走迷宫的入口至出口路径可能不只一条,如何求出所有的路径呢?
求所有路径看起来复杂但其实更简单,只要在老鼠走至出口时显示经过的路径,然后退回上一格重新选择下一个位置继续递归就可以了,比求出单一路径还简单,我們的程序只要作一点修改就可以了。
public class Mouse {
private int startI, startJ; // 入口
private int endI, endJ; // 出口

public static void main(String[] args) {
int maze[][] = {{2, 2, 2, 2, 2, 2, 2, 2, 2},
{2, 0, 0, 0, 0, 0, 0, 0, 2},
{2, 0, 2, 2, 0, 2, 2, 0, 2},
{2, 0, 2, 0, 0, 2, 0, 0, 2},
{2, 0, 2, 0, 2, 0, 2, 0, 2},
{2, 0, 0, 0, 0, 0, 2, 0, 2},
{2, 2, 0, 2, 2, 0, 2, 2, 2},
{2, 0, 0, 0, 0, 0, 0, 0, 2},
{2, 2, 2, 2, 2, 2, 2, 2, 2}};

System.out.println("显示迷宫:");
for(int i = 0; i < maze.length; i++) {
for(int j = 0; j < maze[0].length; j++)
if(maze[i][j] == 2)
System.out.print("■");
else
System.out.print(" ");
System.out.println();
}

Mouse mouse = new Mouse();
mouse.setStart(1, 1);
mouse.setEnd(7, 7);

mouse.go(maze);
}

public void setStart(int i, int j) {
this.startI = i;
this.startJ = j;
}

public void setEnd(int i, int j) {
this.endI = i;
this.endJ = j;
}

public void go(int[][] maze) {
visit(maze, startI, startJ);
}

private void visit(int[][] maze, int i, int j) {
maze[i][j] = 1;

if(i == endI && j == endJ) {
System.out.println("\n找到出口!");
for(int m = 0; m < maze.length; m++) {
for(int n = 0; n < maze[0].length; n++) {
if(maze[m][n] == 2)
System.out.print("■");
else if(maze[m][n] == 1)
System.out.print("◇");
else
System.out.print(" ");
}
System.out.println();
}
}

if(maze[i][j+1] == 0)
visit(maze, i, j+1);
if(maze[i+1][j] == 0)
visit(maze, i+1, j);
if(maze[i][j-1] == 0)
visit(maze, i, j-1);
if(maze[i-1][j] == 0)
visit(maze, i-1, j);

maze[i][j] = 0;
}
}
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