本文整理汇总了Python中region.Region.remove方法的典型用法代码示例。如果您正苦于以下问题：Python Region.remove方法的具体用法？Python Region.remove怎么用？Python Region.remove使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类region.Region的用法示例。
在下文中一共展示了Region.remove方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: find_path
# 需要导入模块: from region import Region [as 别名]
# 或者: from region.Region import remove [as 别名]
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