大佬教程收集整理的这篇文章主要介绍了试图数数的错误。从网格上的 A 点到 B 点所需的移动数。不适用于所有值，大佬教程大佬觉得挺不错的，现在分享给大家，也给大家做个参考。
我正在挑战计算从 A 点到 B 点所需的移动次数，该网格像棋盘一样布置，您可以进行的移动是骑士的移动，因此 2 英寸任意方向和1个垂直。
我已经解决了大部分问题，但由于某种原因，我的计数器没有返回两点之间的移动次数。以下是我对计数的看法。
您会注意到我使用了一个名为 position 的字典，这样做的原因是我可以存储一个 int 值，表示该特定位置从目的地开始的移动次数。
我想最后我应该在移动被认为有效后增加移动值，但我仍然没有得到正确的数字。
def solution(src,dest):
# Chessboard made using nested Lists. The indexes will act as coordinates.
chessboard = [
[0,1,2,3,4,5,6,7],[8,9,10,11,12,13,14,15],[16,17,18,19,20,21,22,23],[24,25,26,27,28,29,30,31],[32,33,34,35,36,37,38,39],[40,41,42,43,44,45,46,47],[48,49,50,51,52,53,54,55],[56,57,58,59,60,61,62,63]
]
# Find index values of src and dest
for row in chessboard:
if src in row:
srcX = chessboard.index(row)
srcY = row.index(srC)
if Dest in row:
destX = chessboard.index(row)
destY = row.index(dest)
# position Dict to store indexes and no of mvoes when using bfs
position = {
'x': 0,'y': 0,'moves': 0,}
position['x'] = srcX
position['y'] = srcY
# Below represents the knights moves to be applIEd to the index of position
row = [-2,-2,-1,-1]
col = [-1,-2]
# We use an if-statement to check for valID moves
def isValID(x,y):
return not (x < 0 or y < 0 or x >=8 or y >=8)
q = []
q.append(position)
# Record spaces visited already
isVisited = []
while len(q)>0:
space = q.pop()
x = space['x']
y = space['y']
moves = space['moves']
# if the position matches the desTination,return no.moves
# I'm just using print to see the result in the terminal
if x == destX and y == destY:
print(moves)
if (x,y) not in isVisited:
isVisited.append((x,y))
# Loop over possible moves
for i in range(len(row)):
newX = x + row[i]
newY = y + col[i]
if isValID(newX,newY):
position['x'] = newX
position['y'] = newY
position['moves'] = moves+1
q.append(position)
这是一个常见的 Python 问题。您创建一个名为“位置”的字典并将其添加到队列中。然后，每当您有新动作时，您都可以修改同一个字典并将其再次添加到队列中，但这并不会创建新字典。你最终会得到一个队列，里面充满了对完全相同的字典的引用。每次更改“位置”时，都会更改队列中的每个字典。每次都需要创建一个新对象。
这似乎有效。
# We use an if-statement to check for valid moves
def isValid(x,y):
return (0 <= x <= 7) and (0 <= y <= 7)
def solution(src,dest):
# Chessboard made using nested lists. The indexes will act as coordinates.
chessboard = [
[0,1,2,3,4,5,6,7],[8,9,10,11,12,13,14,15],[16,17,18,19,20,21,22,23],[24,25,26,27,28,29,30,31],[32,33,34,35,36,37,38,39],[40,41,42,43,44,45,46,47],[48,49,50,51,52,53,54,55],[56,57,58,59,60,61,62,63]
]
# Find index values of src and dest
srcY = src % 8
srcX = src // 8
destY = dest % 8
destX = dest // 8
# Position Dict to store indexes and no of mvoes when using bfs
position = {
'x': srcX,'y': srcY,'moves': 0,}
# Below represents the knights moves to be applied to the index of position
row = [ -2,-2,-1,2]
col = [ -1,1]
q = []
q.append(position)
# Record spaces visited already
isVisited = []
while q:
space = q.pop()
print( "checking",space )
x = space['x']
y = space['y']
moves = space['moves']
# if the position matches the desTination,return no.moves
# I'm just using print to see the result in the terminal
if x == destX and y == destY:
print(f"{moves}!!!")
return moves
if (x,y) not in isVisited:
isVisited.append((x,y))
# Loop over possible moves
for dx,dy in zip(row,col):
newX = x + dx
newY = y + dy
if isValid(newX,newY):
position = {
'x': newX,'y': newY,'moves': moves+1
}
q.append(position)
print( solution( 3,61 ) )
以上是大佬教程为你收集整理的试图数数的错误。从网格上的 A 点到 B 点所需的移动数。不适用于所有值全部内容，希望文章能够帮你解决试图数数的错误。从网格上的 A 点到 B 点所需的移动数。不适用于所有值所遇到的程序开发问题。
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