假設：老鼠在一個二維地圖中i行走，地圖中大部分路徑被墻阻斷，無法前進。老鼠可以按照嘗試錯誤的方法找到出口。只是，這只老鼠必須具備走錯路時候就退回來，并且把走過的路記下來，避免下次走重復路，知道找到出口。

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1.一次只能走一格。

2.遇到墻飾無法前進，退回，找其他方向的路，看是否可以走通。

3.走過的路不會在走第二次。

先利用二維數組MAZE[row][col]生成仿真迷宮，

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?MAZE[i][j] = 1 ,表示[i][j]位置處有墻，無法通過

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??MAZE[i][j] = 10,表示[i][j]位置處無墻，可以通過

MAZE[1][1] 為入口，MAZE[m][n]為出口，迷宮地圖如下

int MAZE[10][12] = { 1,1,1,1,1,1,1,1,1,1,1,1,?? ??? ?//聲明迷宮數組
?? ??? ??? ??? ??? ? 1,0,0,0,1,1,1,1,1,1,1,1,
?? ??? ??? ??? ??? ? 1,1,1,0,1,1,0,0,0,0,1,1,
?? ??? ??? ??? ??? ? 1,1,1,0,1,1,0,1,1,0,1,1,
?? ??? ??? ??? ??? ? 1,1,1,0,0,0,0,1,1,0,1,1,
?? ??? ??? ??? ??? ? 1,1,1,0,1,1,0,1,1,0,1,1,
?? ??? ??? ??? ??? ? 1,1,1,0,1,1,0,1,1,0,1,1,
?? ??? ??? ??? ??? ? 1,1,1,1,1,1,0,1,1,0,1,1,
?? ??? ??? ??? ??? ? 1,1,0,0,0,0,0,0,1,0,0,1,
?? ??? ??? ??? ??? ? 1,1,1,1,1,1,1,1,1,1,1,1 };

//int main()
//{
//    std::cout << "Hello World!\n";
//}
#include <iostream>
#define EAST  MAZE[x][y+1]	//定義東方的相對位置
#define WEST  MAZE[x][y-1]	//定義西方的相對位置
#define SOUTH MAZE[x+1][y]	//定義南方的相對位置
#define NORTH MAZE[x-1][y]	//定義北方的相對位置
using namespace std;
const int ExitX = 8;		//定義出口的X坐標在第8行
const int ExitY = 10;		//定義出口的Y坐標在第10列
struct list
{int x, y;struct list* next;
};   
typedef struct list node;
typedef node* link;
int MAZE[10][12] = { 1,1,1,1,1,1,1,1,1,1,1,1,		//聲明迷宮數組1,0,0,0,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,0,1,1,1,1,1,0,1,1,0,1,1,0,1,1,1,1,1,0,0,0,0,1,1,0,1,1,1,1,1,0,1,1,0,1,1,0,1,1,1,1,1,0,1,1,0,1,1,0,1,1,1,1,1,1,1,1,0,1,1,0,1,1,1,1,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1 };
link push(link stack, int x, int y);
link pop(link stack, int* x, int* y);
int chkExit(int, int, int, int);
int main(void)
{int i, j;link path = NULL;int x = 1;		//入口的X坐標int y = 1;    	//入口的Y坐標cout << "[迷宮的路徑(0的部分)]" << endl; //打印出迷宮的路徑圖for (i = 0; i < 10; i++){for (j = 0; j < 12; j++)cout << MAZE[i][j] << " ";cout << endl;}while (x <= ExitX && y <= ExitY){MAZE[x][y] = 2;if (NORTH == 0){x -= 1;path = push(path, x, y);}else if (SOUTH == 0){x += 1;path = push(path, x, y);}else if (WEST == 0){y -= 1;path = push(path, x, y);}else if (EAST == 0){y += 1;path = push(path, x, y);}else if (chkExit(x, y, ExitX, ExitY) == 1)	//檢查是否走到出口了break;else{MAZE[x][y] = 2;path = pop(path, &x, &y);}}cout << "[老鼠走過的路徑(2的部分)]" << endl; //打印出老鼠成功走出迷宮的路徑圖for (i = 0; i < 10; i++){for (j = 0; j < 12; j++)cout << MAZE[i][j] << " ";cout << endl;}system("pause");return 0;
}
link push(link stack, int x, int y)
{link newnode;newnode = new node;if (!newnode){cout << "Error!內存分配失敗!" << endl;return NULL;}newnode->x = x;newnode->y = y;newnode->next = stack;stack = newnode;return stack;
}
link pop(link stack, int* x, int* y)
{link top;if (stack != NULL){top = stack;stack = stack->next;*x = top->x;*y = top->y;delete top;return stack;}else*x = -1;return stack;
}
int chkExit(int x, int y, int ex, int ey)
{if (x == ex && y == ey){if (NORTH == 1 || SOUTH == 1 || WEST == 1 || EAST == 2)return 1;if (NORTH == 1 || SOUTH == 1 || WEST == 2 || EAST == 1)return 1;if (NORTH == 1 || SOUTH == 2 || WEST == 1 || EAST == 1)return 1;if (NORTH == 2 || SOUTH == 1 || WEST == 1 || EAST == 1)return 1;}return 0;
}

?

#include <iostream>//
//連接圖像中斷裂的邊緣
//以某一點(i,j)為中心,分析它的八鄰域
//int ConnectEdge(IplImage* src)
{if (NULL == src)return 1;int width = src->width;int height = src->height;uchar* data = (uchar*)src->imageData;for (int i = 2; i < height - 2; i++){for (int j = 2; j < width - 2; j++){//如果該中心點為255,則考慮它的八鄰域if (data[i * src->widthStep + j] == 255){int num = 0;for (int k = -1; k < 2; k++){for (int l = -1; l < 2; l++){//如果八鄰域中有灰度值為0的點，則去找該點的十六鄰域if (k != 0 && l != 0 && data[(i + k) * src->widthStep + j + l] == 255)num++;}}//如果八鄰域中只有一個點是255，說明該中心點為端點，則考慮他的十六鄰域if (num == 1){for (int k = -2; k < 3; k++){for (int l = -2; l < 3; l++){//如果該點的十六鄰域中有255的點，則該點與中心點之間的點置為255if (!(k < 2 && k > -2 && l < 2 && l > -2) && data[(i + k) * src->widthStep + j + l] == 255){data[(i + k / 2) * src->widthStep + j + l / 2] = 255;}}}}}}}
}

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