本文整理匯總了Python中region.Region.remove方法的典型用法代碼示例。如果您正苦於以下問題：Python Region.remove方法的具體用法？Python Region.remove怎麽用？Python Region.remove使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類region.Region的用法示例。
在下文中一共展示了Region.remove方法的1個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的Python代碼示例。
示例1: find_path
# 需要導入模塊: from region import Region [as 別名]
# 或者: from region.Region import remove [as 別名]
def find_path(self, loc1, loc2, pid=None, cPane=None):
'''
Pathfinding algorithm using a tweaked implementation of Dijkstra's Algorithm
The difference is that we may not need a path directly to loc2. If pid refers to a valid
Person object, we just need a path to the closest location that is within attack range of
the final target. So loop through all possible locations searching for the location with the
shortest path, and return that.
'''
if loc1 == loc2:
return []
if loc1.distance(loc2) == 1:
return [loc2]
Q = Region("SQUARE", Location(loc1.pane, (0, 0)), Location(loc1.pane, (PANE_X - 1, PANE_Y - 1))).locations
dist = {}
prev = {}
q = PriorityQueue()
for v in Q:
dist[v] = float('inf')
prev[v] = None
dist[loc1] = 0
q.put((0, loc1))
while not q.empty():
u = q.get()[1]
if u == loc2 or dist[u] == float('inf'):
break
neighbors = [u.move(dir) for dir in [2, 4, 6, 8]]
neighbors = [v for v in neighbors if v in Q and (self.tile_is_open(v, pid, cPane) or v == loc2)]
for v in neighbors:
Q.remove(v)
alt = dist[u] + u.distance(v)
if alt < dist[v]:
dist[v] = alt
prev[v] = u
q.put((dist[v], v))
path = []
u = loc2
# Update destination to a location within attack range of the ultimate target
if pid:
minDist = (dist[u], u)
for loc in [l for l in Region("DIAMOND", u, self.person[pid].attackRange - 1) if l.pane == loc1.pane]:
if prev[loc] and dist[loc] < minDist[0]:
minDist = (dist[loc], loc)
u = minDist[1]
# If still no valid destination, find the closest point in anticipation of an obstacle
# eventually being removed, such as an ally getting killed
radius = 1
while not prev[u] and radius < 10:
R = Region("SQUARE", loc2, radius) - Region("SQUARE", loc2, radius - 1)
for loc in [l for l in R if l.pane == loc1.pane]:
if prev[loc]:
u = loc
break
radius += 1
# Follow the path backwards from destination to reconstitute the shortest path
while prev[u]:
path.append(u)
u = prev[u]
path.reverse()
return path