说明

老鼠走迷宫是递回求解的基本题型.我们在二维阵列中使用 2 表示迷宫墙壁,使用 1 来表示老鼠的行走路径,尝试以程式求由入口至出口的路径.

解法

老鼠可走上,下,左,右四个方向.入口一般在左上,出口在右下(如果说有其他出入口情况则下述老鼠每步默认选择方向的顺序改变)

老鼠从入口出发,每走一步进行一次四个方向的选择,判断下步是否有可走方向(由于入口在左上,出口在左下,则优先选择右,下方向,之后选择左,上方向).无法前进时退回上步选择下一个可前进方向.

如此在阵列中按照顺序测试四个方向,直到走出出口为止.

代码(c语言版,思路可参考)

参数声明

int maze[7][7] = { {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},

{2,2,0,2,0,2,0},

{2,0,0,0,0,0,2},

{2,2,2,2,2,2,2} };//mazing map

int(*mazeP)[7];

mazeP = maze;//将二维数组传递给数组指针(数组指针是指针指向一个数组)

int startI = 1, startJ = 1, endI = 5, endJ = 5;//entrance and exit

int success = 0;//success mark

int i = 1,j = 1;//initialization of import and export

复制代码递回方法

//递回进行老鼠每步选择的方向,返回是否成功

int mazing_visit(int i, int j,int(*mazeP)[7],int success)

{

mazeP[i][j] = 1;

if (i == 5 && j == 5)success = 1;

//优先右下,然后左上

if ( 1 != success && mazeP[i][j + 1] == 0)mazing_visit(i, j + 1, mazeP, success);

if (success != 1 && mazeP[i + 1][j] == 0)mazing_visit(i + 1, j, mazeP, success);

if (success != 1 && mazeP[i][j - 1] == 0)mazing_visit(i, j - 1, mazeP, success);

if (success != 1 && mazeP[i - 1][j] == 0)mazing_visit(i - 1, j, mazeP, success);

if (success != 1)mazeP[i][j] == 0;

return success;

}

复制代码完整代码

#include

#include

#include

//Mazing-老鼠走迷宫

int mazing_visit(int, int,int (*p)[7],int);

void mazing()

{

int maze[7][7] = { {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},

{2,2,0,2,0,2,0},

{2,0,0,0,0,0,2},

{2,2,2,2,2,2,2} };//mazing map

int(*mazeP)[7];

mazeP = maze;//将二维数组传递给数组指针(数组指针是指针指向一个数组)

int startI = 1, startJ = 1, endI = 5, endJ = 5;//entrance and exit

int success = 0;//success mark

int i = 1,j = 1;//initialization of import and export

printf("显示迷宫: \n");

for (i = 0; i < 7; i++)

{

for (j = 0; j < 7; j++)

{

if (mazeP[i][j] == 2)//判断是否是墙壁

{

printf("*");

}

else

{

printf(" ");

}

}

printf("\n");

}

//判断是否存在出口,若有出口则打印出路径

if (mazing_visit(startI, startJ, mazeP,success) == 0)

{

printf("\n没有找到出口\n");

}

else

{

printf("\n显示路径:");

for (i = 0; i < 7; i++)

{

for (j = 0; j < 7; j++)

{

if (mazeP[i][j] == 2)

{

printf("*");

}

else if (mazeP[i][j] == 1)

{

printf(">");

}

else

{

printf(" ");

}

}

printf("\n");

}

}

}

int mazing_visit(int i, int j,int(*mazeP)[7],int success)

{

mazeP[i][j] = 1;

if (i == 5 && j == 5)success = 1;

//优先右下,然后左上

if ( 1 != success && mazeP[i][j + 1] == 0)mazing_visit(i, j + 1, mazeP, success);

if (success != 1 && mazeP[i + 1][j] == 0)mazing_visit(i + 1, j, mazeP, success);

if (success != 1 && mazeP[i][j - 1] == 0)mazing_visit(i, j - 1, mazeP, success);

if (success != 1 && mazeP[i - 1][j] == 0)mazing_visit(i - 1, j, mazeP, success);

if (success != 1)mazeP[i][j] == 0;

return success;

}

int main(void)

{

mazing();

return 0;

}

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