import Astar import heapq start_cor = (19, 0) waypoints = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) def distance(_from, _to): x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(waypoints) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(waypoints[i], waypoints[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist start = 0 end = n - 1 distances = [[float('inf')] * (n + 1) for _ in range(n)] visited = set() heap = [(0, 0, start)] while heap: (dist, num_visited, current) = heapq.heappop(heap) if current == end and num_visited == 8: break if (current, num_visited) in visited: continue visited.add((current, num_visited)) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0: new_num_visited = num_visited if neighbor in range(start + 1, end) and (current not in range(start + 1, end)) and num_visited < 8: new_num_visited += 1 new_distance = dist + weight if new_distance < distances[neighbor][new_num_visited]: distances[neighbor][new_num_visited] = new_distance heapq.heappush(heap, (new_distance, new_num_visited, neighbor)) min_dist = float('inf') min_num_visited = 8 for i in range(8): if distances[end][i] < min_dist: min_dist = distances[end][i] min_num_visited = i path = [end] current = end num_visited = min_num_visited for i in range(len(waypoints), 0, -1): if current in range(i): num_visited -= 1 for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and (neighbor, num_visited) in visited and distances[neighbor][num_visited] + weight == \ distances[current][num_visited]: path.append(neighbor) current = neighbor break path.reverse() print(f"The optimal path from start to end through the 8 waypoints is: {path}") print(f"The total distance is: {distances[end][min_num_visited]}")

这段代码是用来解决一个路径规划问题，目标是找到从起点到终点经过8个给定点的最短路径。其中使用了A*算法来计算两点之间的距离，并利用堆来实现最小优先队列。具体流程是先计算出8个给定点之间的距离，再从起点开始，每次取出离当前点最近的点，如果该点是终点并且已经经过了8个给定点，则找到了最短路径。如果该点不是终点，则将该点的所有邻居加入队列中，并更新到达邻居的距离和经过的给定点数。最后根据经过的给定点数从终点开始倒推出最短路径。

相关问题

import Astar import heapq start_cor = (19, 0) waypoints = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) # Define a function to calculate the distance between two coordinates def distance(_from, _to): x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath # Compute the distances between all pairs of waypoints n = len(waypoints) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(waypoints[i], waypoints[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist # Set the start and end points start = 0 end = n - 1 # Initialize the distances and visited set distances = [[float('inf')] * (n+1) for _ in range(n)] visited = set() # Initialize the heap with the start point and 0 distance and 0 waypoints visited heap = [(0, 0, start)]

这段代码使用A*算法和堆（heap）数据结构实现了从起点到终点经过所有指定航点的最短路径。首先，定义了起点、终点和指定航点的坐标。然后，定义了一个计算两个坐标之间距离的函数distance，内部使用了A*算法计算两个坐标之间的最短路径长度。接着，使用上述计算函数计算了所有航点之间的距离，并存储在邻接矩阵adj_matrix中。接下来，定义了起点、终点到所有航点的距离distances和已经访问过的航点集合visited，并初始化堆heap，将起点和0距离以及0已访问航点数加入堆中。最后，可以使用堆来实现Dijkstra算法，找到从起点到终点经过所有指定航点的最短路径。

import Astar import heapq start_cor = (19, 0) treasures = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end_cor = (1, 20) # 定义一个函数计算两个坐标之间的距离 def distance(_from, _to): # 返回从起点到终点的最短路径 x1, y1 = _from x2, y2 = _to distancepath = Astar.find_path(x1, y1, x2, y2) return distancepath n = len(treasures) adj_matrix = [[0] * n for _ in range(n)] for i in range(n): for j in range(i + 1, n): dist = distance(treasures[i], treasures[j]) adj_matrix[i][j] = dist adj_matrix[j][i] = dist # 使用Dijkstra算法求解最短路径 start = 0 end = n - 1 distances = [float('inf')] * n distances[start] = 0 visited = set() heap = [(0, start)] while heap: (dist, current) = heapq.heappop(heap) if current == end: break if current in visited: continue visited.add(current) for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and neighbor not in visited: new_distance = dist + weight if new_distance < distances[neighbor]: distances[neighbor] = new_distance heapq.heappush(heap, (new_distance, neighbor)) # 输出结果 path = [end] current = end while current != start: for neighbor, weight in enumerate(adj_matrix[current]): if weight > 0 and distances[current] == distances[neighbor] + weight: path.append(neighbor) current = neighbor break print(path) path.reverse() print(f"从第{start+1}个坐标开始经过其他几个坐标最后到达第{end+1}个坐标的最短路线为：{path}") print(f"总距离为：{distances[end]}"

这段代码实现的是与上一段代码几乎相同的功能，不同之处在于使用了不同的变量名和起点终点坐标的表示方式。具体来说，代码完成以下步骤：

1. 导入 Astar 模块和 heapq 模块。
2. 定义起点坐标 `start_cor` 和终点坐标 `end_cor`，以及一个包含多个宝藏坐标的列表 `treasures`。
3. 定义一个函数 `distance`，用于计算两个坐标之间的距离。在该函数中，调用了 A* 算法的 `find_path` 函数来计算从起点到终点的最短路径，并将路径长度作为距离返回。
4. 根据宝藏坐标列表 `treasures`，使用 `distance` 函数计算出各个宝藏之间的距离，构建邻接矩阵 `adj_matrix`。
5. 使用 Dijkstra 算法求解从起点到终点的最短路径。在该算法中，使用 `heapq` 模块维护一个优先队列，每次从队列中取出距离最小的节点进行遍历，并更新其相邻节点的距离。
6. 根据 Dijkstra 算法的结果，使用邻接矩阵和起点终点信息，回溯出从起点到终点的最短路径，并输出路径信息和总距离。

相对于上一段代码，这段代码使用了更加直观的变量名，同时在计算距离时也更加直接，直接使用 A* 算法计算最短路径的长度。