用于查找与给定终点、起点和最小转弯数匹配的直线的程序

import heapq

def heuristic(a, b):
    return abs(a[0] - b[0]) + abs(a[1] - b[1])

def a_star_search(graph, start, goal, min_turns):
    frontier = []
    heapq.heappush(frontier, (0, start))
    came_from = {start: None}
    cost_so_far = {start: 0}
    turns = {start: 0}

    while frontier:
        current = heapq.heappop(frontier)[1]

        if current == goal and turns[current] >= min_turns:
            break

        for next_node in graph.neighbors(current):
            new_cost = cost_so_far[current] + graph.cost(current, next_node)
            new_turns = turns[current] + (1 if came_from[current] != next_node else 0)

            if next_node not in cost_so_far or new_cost < cost_so_far[next_node]:
                cost_so_far[next_node] = new_cost
                priority = new_cost + heuristic(goal, next_node)
                heapq.heappush(frontier, (priority, next_node))
                came_from[next_node] = current
                turns[next_node] = new_turns

    if goal not in came_from:
        return None, None

    path = []
    current = goal
    while current != start:
        path.append(current)
        current = came_from[current]
    path.append(start)
    path.reverse()

    return path, turns[goal]

# Example usage
class Graph:
    def __init__(self):
        self.edges = {}
    
    def neighbors(self, node):
        return self.edges.get(node, [])
    
    def cost(self, current, next_node):
        return 1  # Assuming all edges have the same cost for simplicity

graph = Graph()
# Add edges to the graph
# graph.edges = {...}

start = (0, 0)
goal = (4, 4)
min_turns = 2

path, turns = a_star_search(graph, start, goal, min_turns)
print("Path:", path)
print("Turns:", turns)