少种走法，打印每一种走法。

/*==========================================================*\ 给二维数组，存（0，0） -> (m, n)个点，就是一个m * n 的网格， 从左上角（0,0）走到右下角（m，n）只能向右走或者向下走，有多 少种走法，打印每一种走法。\*==========================================================*/#include 
    
     #include 
     
      using namespace std; struct position{	int x;	int y;};position *stepX(position *pCurrent){	position *newStep = new position();	newStep->x = pCurrent->x + 1;	newStep->y = pCurrent->y;	return newStep;}position *stepY(position *pCurrent){	position *newStep = new position();	newStep->x = pCurrent->x;	newStep->y = pCurrent->y + 1;	return newStep;}void path(int widthM,int heightN,deque
      
        &dq){	int endM = widthM - 1;	int endN = heightN - 1;	if(dq.back()->x == endM && dq.back()->y == endN){		deque
       
        ::iterator it = dq.begin();		while (it != dq.end()){			cout << "[" << (*it)->x << "," << (*it)->y << "]" << " ";			it++;		}		cout << endl;		return ;	}else if(dq.back()->y > endN){		return ;	}else if(dq.back()->x > endM){		return ;	}else{		position *newStep = stepX(dq.back());		dq.push_back(newStep);		path(widthM,heightN,dq);		dq.pop_back();		newStep = stepY(dq.back());		dq.push_back(newStep);		path(widthM,heightN,dq);		dq.pop_back();	}}int main(){	deque
        
          dq;	position *newStep = new position;	newStep->x = 0;	newStep->y = 0;	dq.push_back(newStep);	path(2,3,dq);	system("pause");	return 1;}