[英]Finding the shortest path between the two selected cells (If you can not go diagonally)
我有一個矩陣:
maze = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0]]
1-障礙
0-常規單元格
我想實現一種算法,以找到兩個選定像元之間的最短路徑(如果不能對角線的話)。 我嘗試了A *算法,但未給出正確的結果:
def astar(maze, start, end):
start_node = Node(None, start)
start_node.g = start_node.h = start_node.f = 0
end_node = Node(None, end)
end_node.g = end_node.h = end_node.f = 0
open_list = []
closed_list = []
open_list.append(start_node)
while len(open_list) > 0:
current_node = open_list[0]
current_index = 0
for index, item in enumerate(open_list):
if item.f < current_node.f:
current_node = item
current_index = index
open_list.pop(current_index)
closed_list.append(current_node)
if current_node == end_node:
path = []
current = current_node
while current is not None:
path.append(current.position)
current = current.parent
return path[::-1] # Return reversed path
children = []
for new_position in [(0, -1), (0, 1), (-1, 0), (1, 0), (-1, 1), (1, -1)]:
node_position = (current_node.position[0] + new_position[0], current_node.position[1] + new_position[1])
if node_position[0] > (len(maze) - 1) or node_position[0] < 0 or node_position[1] > (len(maze[len(maze)-1]) -1) or node_position[1] < 0:
continue
if maze[node_position[0]][node_position[1]] != 0:
continue
new_node = Node(current_node, node_position)
children.append(new_node)
for child in children:
for closed_child in closed_list:
if child == closed_child:
continue
child.g = current_node.g + 1
child.h = ((child.position[0] - end_node.position[0]) ** 2) + ((child.position[1] - end_node.position[1]) ** 2)
child.f = child.g + child.h
for open_node in open_list:
if child == open_node and child.g > open_node.g:
continue
open_list.append(child)
請告訴我如何用Python語言實現它,以使其正常工作。
這是針對您問題的BFS實現: https : //ideone.com/tuBu3G我們從感興趣的起點開始我們的隊列,並在訪問終點后停止。 在迭代的每一步中,我們的目標都是探索新的未探索狀態,並將此新點的距離設置為1 +該點到其探索位置的距離 。
from collections import deque
graph = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0]]
# To move left, right, up and down
delta_x = [-1, 1, 0, 0]
delta_y = [0, 0, 1, -1]
def valid(x, y):
if x < 0 or x >= len(graph) or y < 0 or y >= len(graph[x]):
return False
return (graph[x][y] != 1)
def solve(start, end):
Q = deque([start])
dist = {start: 0}
while len(Q):
curPoint = Q.popleft()
curDist = dist[curPoint]
if curPoint == end:
return curDist
for dx, dy in zip(delta_x, delta_y):
nextPoint = (curPoint[0] + dx, curPoint[1] + dy)
if not valid(nextPoint[0], nextPoint[1]) or nextPoint in dist.keys():
continue
dist[nextPoint] = curDist + 1
Q.append(nextPoint)
print(solve((0,0), (6,7)))
版畫:#13
這是一個如何在python 3中實現BFS的示例:
import collections
def breadth_first_search(graph, root):
visited, queue = set(), collections.deque([root])
while queue:
vertex = queue.popleft()
for neighbour in graph[vertex]:
if neighbour not in visited:
visited.add(neighbour)
queue.append(neighbour)
if __name__ == '__main__':
graph = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 1, 0, 0, 0, 0, 0]]
breadth_first_search(graph, 0)
希望這能夠對您有所幫助,請讓我知道如何進行!
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