我们使用了A*算法的启发式(曼哈顿距离)来改进Dijkstra算法的性能。当我们将邻居节点添加到优先队列时,我们使用了distance + heuristic作为优先级,这样我们可以更快地找到通往目标节点的路径。
python
import heapq
import numpy as np
def heuristic(a, b):
(x1, y1) = a
(x2, y2) = b
return abs(x1 - x2) + abs(y1 - y2) # 使用曼哈顿距离作为启发式
def dijkstra_with_a_star_heuristic(graph, start, end):
# 初始化距离字典,将所有节点设置为无穷大,除了起点
distances = {position: float('infinity') for position in np.ndindex(graph.shape)}
distances[start] = 0
# 优先队列,存储待检查的节点和它们的距离
pq = [(0, start)]
while pq:
# 弹出当前最小距离的节点
current_distance, current_position = heapq.heappop(pq)
# 如果已经找到更短的路径到当前节点,则跳过
if current_distance > distances[current_position]:
continue
# 如果到达目标节点,返回路径(这里未实现路径重构)
if current_position == end:
return distances[current_position] # 这里只返回距离,路径重构需要额外工作
# 遍历当前节点的邻居
for dx, dy in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
x2, y2 = current_position[0] + dx, current_position[1] + dy
# 检查邻居是否在网格内且不是障碍物
if 0 <= x2 < graph.shape[0] and 0 <= y2 < graph.shape[1] and graph[x2, y2] != 1:
# 计算到达邻居节点的距离
distance = current_distance + 1
# 使用启发式来改进Dijkstra的选择
priority = distance + heuristic((x2, y2), end)
# 如果找到更短的路径到邻居节点,则更新距离并添加到优先队列中
if priority < distances[(x2, y2)]:
distances[(x2, y2)] = priority
heapq.heappush(pq, (priority, (x2, y2)))
# 如果没有找到路径到目标节点
return None
if __name__ == '__main__':
# 示例用法
graph = np.array([
[0, 0, 0, 1, 0],
[1, 1, 0, 1, 0],
[0, 0, 0, 0, 0],
[0, 1, 1, 1, 1],
[0, 0, 0, 0, 0]
])
start = (0, 0)
end = (4, 4)
distance = dijkstra_with_a_star_heuristic(graph, start, end)
print(f"The shortest distance from {start} to {end} is: {distance}")