這是資料結構的作業，便找書邊看網上，然後自己慢慢寫出來的,這裡面主要是回溯法。
因為課本上是打印出一條路徑，然後我在想怎樣能將所有的路徑都輸出來，方法：就是當求出一條路徑後，將出口點變成可以走的點（因為之前將其值變成了-1），並且將棧頂元素出棧，還需要得到現在棧頂元素的i，j，di值，將其賦出來。
這裡的思路是這樣的，因為找最後一個點的時候是找倒數第二個點的上下左右四個方位，假設說是路徑通暢的點是倒數第二個點的上面的點，當我們找另一條路徑時，那麼現在我們應該順時針從次棧頂右面的點試探，檢視是不是可行的點。


#include <stdio.h>
#define Max 9999
int mg[6][6] = {
			  {1,1,1,1,1,1},{1,0,0,0,1,1},
			  {1,0,1,0,0,1},{1,0,0,0,1,1},
			  {1,1,0,0,0,1},{1,1,1,1,1,1}
};
typedef struct {
	int i;
	int j;
	int di;
}BOX;

typedef struct {
	BOX data[Max];
	int top;
}SqStack;

int InitSqStack(SqStack *s)
{
	s->top = -1;
	return 0;
}

int Destroy(SqStack *s)
{
	free(s);
	return 0;
}
int Push(SqStack *s, BOX *e)
{
	if (s->top == Max - 1)
		return 0;
	s->top++;
	s->data[s->top] = *e;
	return 0;
}
int GetTop(SqStack *s, BOX *e)
{
	if (s->top == -1)
		return 1;
	*e = s->data[s->top];
	return 0;
}
int Pop(SqStack *s, BOX *e)
{
	if (s->top == -1)
		return 1;
	*e = s->data[s->top];
	s->top--;
	return 0;
}
int Empty(SqStack *s)
{
	if (s->top == -1)
		return 1;
	else
		return 0;
}//1是空，0是非空。
int Man(SqStack *s)
{
	if (s->top == Max - 1)
		return 1;
	else
		return 0;
}



int zmg(int x, int y, int xe, int ye)
{
	SqStack S;
	BOX path[Max], e, E;
	BOX shortpath[Max];
	int di = 0, x1 = 0, y1 = 0, i = 0, j = 0;
	int k = 0, fun = 0;
	int m = 0, lm = 0, n = 0, op = 0;
	int minlen = Max;
	int js = 0, tp = 0;
	InitSqStack(&S);//初始化棧
	e.i = x;
	e.j = y;
	e.di = -1;
	Push(&S, &e);
	mg[x][y] = -1;
	while (Empty(&S) != 1)//不等於空
	{
		GetTop(&S, &e);
		i = e.i;
		j = e.j;
		di = e.di;
		if (i == xe && j == ye)
		{
			k = 0;
			tp = S.top;
			for (js = 0;js <= tp;js++)
			{
				S.top = js;
				printf("\t(%d,%d)", S.data[S.top].i, S.data[S.top].j);
				shortpath[n] = path[op];
			}
			printf("\n");
			
			if (tp + 1 < minlen)
			{	
				for (S.top = 0,n = 0;n <=tp;n++,S.top++)
				{
					shortpath[n] = S.data[S.top];		
				}
				S.top = tp;
				minlen = tp + 1;
			}
			mg[i][j] = 0;
			Pop(&S, &e);
			GetTop(&S, &e);
			i = e.i;
			j = e.j;
			di = e.di;
		}
		fun = 0;
		while (di < 4 && fun != 1)
		{
			di++;
			switch (di)
			{
			case 0:
				x1 = i - 1;
				y1 = j;
				break;
			case 1:
				x1 = i;
				y1 = j + 1;
				break;
			case 2:
				x1 = i + 1;
				y1 = j;
				break;
			case 3:
				x1 = i;
				y1 = j - 1;
				break;
			}
			if (mg[x1][y1] == 0)
				fun = 1;
		}
		if (fun == 1)
		{
			S.data[S.top].di = di;
			e.i = x1;
			e.j = y1;
			e.di = -1;
			Push(&S, &e);
			mg[x1][y1] = -1;
		}
		else
		{
			Pop(&S, &e);
			mg[e.i][e.j] = 0;
		}
	}
	      printf("\n");
	      printf("最短路徑的長度%d，M:",minlen);
	      for(n=0;n<minlen;n++)
	     {
	             printf("\t(%d,%d)",shortpath[n].i,shortpath[n].j);
	    }
	      Destroy(&S);
	return 0;
}
int main()
{
	printf("迷宮所有路徑如下:\n");
	zmg(1, 1, 4, 4);
	return 0;
}