所以我试图编写一个代码来确定在一个电路板上放置哪些形状,并且如果电路板上已经有一些小块,它们会知道它们是否重叠。我需要Python中的帮助来检查列表的边界
所以我现在有的代码,需要用户输入来获得董事会的长度和。
然后假设用户知道电路板的大小,它会提示您输入电路板上已有的任何零件。
因此,如果板是一个5×4和它有一个L形的,它应该是这样的:
[1,1,0,0,0,
1,0,0,0,0,
1,0,0,0,0,
0,0,0,0,0]
但是,你会进入[1,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0]
然后,它要求你输入在从默认位置开始的形状中,这将是右上角。
当进入形状的板是这样的:
[1, 2, 3, 4, 5,
6, 7, 8, 9, 10,
11,12,13,14,15,
16,17,18,19,20]
因此,一个L形将有默认坐标:
[1,2,6,11]
然后问下一个可用的一点是,所以如果你打算把L添加到已经存在L的板上。下一个可用的点就是3,如下所示:
[1,1,X,0,0,
1,0,0,0,0,
1,0,0,0,0,
0,0,0,0,0]
所以程序的输出形状的新坐标具体做法是:
[3,4,8,13]
但我需要帮助的错误检查溢出。换句话说,如果一块棋子离开棋盘,程序将打印失败。
例如:如果我想在5处放一个L块,坐标会输出[5,6,11,16],这是不对的。
它应该是什么样子:
[0,0,0,0,X,X
0,0,0,0,X,
0,0,0,0,X,
0,0,0,0,0]
但会发生什么:
[0,0,0,0,X,
X,0,0,0,0,
X,0,0,0,0,
X,0,0,0,0]
我怎样才能让如果一块熄灭板其打印失败了吗?我已经这样做了,所以你不能输入负的坐标,并且你不能让一块块比最大的坐标更远。
这里是我的代码:
user1 = input ('Enter the length of the board: ')
#Length of the board
user2 = input ('Enter the width of the board: ')
#Width of the board
user3 = user1 * user2
#Area of the board
board = input ('Enter in the contenets of the board, as a list (i.e. [0,1,....,0,0]): ')
#Use list to enter in contents of board if any
listA = input ('Enter the shape, in terms of default coordinates (i.e. [1,2,6,11]): ')
#Enter in shape of piece you are placing on board
if listA[-1] > user3:
print ('Failed!')
#Piece fails if end goes off board
else:
if listA[0] < 1:
print ('Failed!')
#Piece fails if begining goes off board
else:
c = int(input ('Enter the next free slot: '))
#Enter where the piece fits next
a = int(listA[0])
#First coordinate of piece
b = [((x + c) - a) for x in listA]
#Finds location of moved piece
overlap = False
#Boolean to check for piece overlap
for y in b:
if board[y - 1] == 1:
overlap = True
#Overlap is true if piece ovelaps
#another piece already on the board
else:
overlap = overlap
#If no overlapping occurs then continue
if overlap == True:
print ('Failed!')
#Fails if piece overlaps
else:
print b
#prints the coordinates of the moved piece
您是否考虑用实际的二维结构(即列表列表)来表示二维“板”? – 2012-02-14 02:24:33
这闻起来像功课。如果是这样,你应该使用'homework'标签。 – millimoose 2012-02-14 02:25:07
@Karl这可能不会真的帮助在这种情况下。你仍然需要明确检查形状的大小,它是否与列表中的len或者存储在变量中的宽度无关。 – millimoose 2012-02-14 02:27:01