所以我試圖編寫一個代碼來確定在一個電路板上放置哪些形狀,並且如果電路板上已經有一些小塊,它們會知道它們是否重疊。我需要Python中的幫助來檢查列表的邊界
所以我現在有的代碼,需要用戶輸入來獲得董事會的長度和。
然後假設用戶知道電路板的大小,它會提示您輸入電路板上已有的任何零件。
因此,如果板是一個5×4和它有一個L形的,它應該是這樣的:
[1,1,0,0,0,
1,0,0,0,0,
1,0,0,0,0,
0,0,0,0,0]
但是,你會進入[1,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0]
然後,它要求你輸入在從默認位置開始的形狀中,這將是右上角。
當進入形狀的板是這樣的:
[1, 2, 3, 4, 5,
6, 7, 8, 9, 10,
11,12,13,14,15,
16,17,18,19,20]
因此,一個L形將有默認座標:
[1,2,6,11]
然後問下一個可用的一點是,所以如果你打算把L添加到已經存在L的板上。下一個可用的點就是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]
所以程序的輸出形狀的新座標具體做法是:
[3,4,8,13]
但我需要幫助的錯誤檢查溢出。換句話說,如果一塊棋子離開棋盤,程序將打印失敗。
例如:如果我想在5處放一個L塊,座標會輸出[5,6,11,16],這是不對的。
它應該是什麼樣子:
[0,0,0,0,X,X
0,0,0,0,X,
0,0,0,0,X,
0,0,0,0,0]
但會發生什麼:
[0,0,0,0,X,
X,0,0,0,0,
X,0,0,0,0,
X,0,0,0,0]
我怎樣才能讓如果一塊熄滅板其打印失敗了嗎?我已經這樣做了,所以你不能輸入負的座標,並且你不能讓一塊塊比最大的座標更遠。
這裏是我的代碼:
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
您是否考慮用實際的二維結構(即列表列表)來表示二維「板」? – 2012-02-14 02:24:33
這聞起來像功課。如果是這樣,你應該使用'homework'標籤。 – millimoose 2012-02-14 02:25:07
@Karl這可能不會真的幫助在這種情況下。你仍然需要明確檢查形狀的大小,它是否與列表中的len或者存儲在變量中的寬度無關。 – millimoose 2012-02-14 02:27:01