A-star搜索(A* search)是一种启发式搜索算法,用于解决问题的最短路径搜索和图搜索问题。它在图中使用了一种启发式评估函数,以估计从当前状态到目标状态的代价。
滑动拼图问题是一个经典的问题,目标是将拼图从初始状态移动到目标状态。下面是使用A-star搜索算法解决滑动拼图问题的代码示例:
import heapq
# 定义滑动拼图问题类
class Puzzle:
def __init__(self, start, goal):
self.start = start
self.goal = goal
# 定义启发式评估函数
def heuristic(self, node):
# 计算当前状态与目标状态之间的曼哈顿距离
distance = 0
for i in range(len(node)):
x1, y1 = divmod(node.index(i), 3)
x2, y2 = divmod(self.goal.index(i), 3)
distance += abs(x1 - x2) + abs(y1 - y2)
return distance
# 定义A-star搜索算法
def astar_search(self):
# 定义优先队列
open_list = []
# 将起始状态加入优先队列中,使用启发式估值作为优先级
heapq.heappush(open_list, (self.heuristic(self.start), self.start))
# 定义已访问状态集合
closed_list = set()
# 定义父状态字典,用于记录每个状态的父状态
parents = {}
# 定义代价字典,用于记录每个状态的代价
costs = {self.start: 0}
while open_list:
# 从优先队列中取出优先级最高的状态
current = heapq.heappop(open_list)[1]
# 如果当前状态为目标状态,停止搜索
if current == self.goal:
break
# 将当前状态加入已访问状态集合中
closed_list.add(current)
# 获取当前状态的邻居状态
neighbors = self.get_neighbors(current)
for neighbor in neighbors:
# 计算邻居状态的代价
new_cost = costs[current] + 1
# 如果邻居状态未被访问过或新的代价更小
if neighbor not in costs or new_cost < costs[neighbor]:
# 更新邻居状态的代价
costs[neighbor] = new_cost
# 更新邻居状态的父状态
parents[neighbor] = current
# 将邻居状态加入优先队列中,使用启发式估值和代价之和作为优先级
heapq.heappush(open_list, (new_cost + self.heuristic(neighbor), neighbor))
# 从目标状态开始,依次追溯父状态,得到最短路径
path = []
current = self.goal
while current != self.start:
path.append(current)
current = parents[current]
path.append(self.start)
path.reverse()
return path
# 获取当前状态的邻居状态
def get_neighbors(self, state):
neighbors = []
# 获取空格的位置
blank_index = state.index(0)
# 上移
if blank_index >= 3:
new_state = state[:]
new_state[blank_index], new_state[blank_index - 3] = new_state[blank_index - 3], new_state[blank_index]
neighbors.append(new_state)
# 下移
if blank_index < 6:
new_state = state[:]
new_state[blank_index], new_state[blank_index + 3] = new_state[blank_index + 3], new_state[blank_index]
neighbors.append(new_state)
# 左移
if blank_index % 3 != 0:
new_state = state[:]
new_state[blank_index], new_state[blank_index - 1] = new_state[blank_index - 1], new_state[blank_index]
neighbors.append(new_state)
#