对以下代码进行逐行解释 if astar == false while (~all(unvisited(:)==inf)) dist_unvisited = dist.*unvisited; [~, id] = min(dist_unvisited(:)); unvisited(id) = inf; num_expanded = num_expanded + 1; [i,j,k] = ind2sub([map.nx,map.ny,map.nz],id); for d = 1:size(dijk,1) nijk = bsxfun(@plus, int32([i,j,k]), int32(dijk(d,:))); if all(nijk > 0) & all(int32([map.nx, map.ny, map.nz]) >= int32(nijk)) ... & unvisited(nijk(1),nijk(2),nijk(3)) == 1 ... & map.occ_map(nijk(1),nijk(2),nijk(3)) ~= 1 alt = dist(id) + sqrt(sum(dijk(d,:).^2)); nid = (nijk(3)-1)*map.nx*map.ny + (nijk(2)-1)*map.nx + nijk(1); if alt < dist(nid) dist(nid) = alt; prev(nijk(1), nijk(2), nijk(3),:) = [i,j,k]; end end end if id == (goal(3)-1)*map.nx*map.ny + (goal(2)-1)*map.nx + goal(1); break end end

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每排是什么意思

这段代码实现了一个寻找最短路径的算法，其中： - waypoints 是一个包含多个点坐标的列表，代表路径上的多个路标点。 - end_cor 是路径的终点坐标。 - distance 函数用于计算两个点之间的距离，使用了 Astar 算法...

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每排是什么意思帮我加上注释

以下是代码的注释解释： # 定义起点和终点的坐标 start_cor = (19, 0) end_cor = (1, 20) # 定义计算两个点之间距离的函数 def distance(_from, _to): x1, y1 = _from x2, y2 = _to # 使用 Astar 算法寻路...

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...

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]}"

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...

import Astar import heapq start = (19, 0) treasures = [(5, 15), (5, 1), (9, 3), (11, 17), (7, 19), (15, 19), (13, 1), (15, 5)] end = (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]}")

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 start 和终点 end 的坐标，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...

OPTIMAL = True CONFIGS = [ (175, ["--search", "astar(merge_and_shrink(merge_strategy=merge_precomputed(merge_tree=linear(variable_order=reverse_level))," "shrink_strategy=shrink_bisimulation(greedy=true)," "label_reduction=exact(before_shrinking=true,before_merging=false)," "max_states=infinity,threshold_before_merge=1))"]), (432, ["--search", "astar(merge_and_shrink(merge_strategy=merge_precomputed(merge_tree=linear(variable_order=reverse_level))," "shrink_strategy=shrink_bisimulation(greedy=false)," "label_reduction=exact(before_shrinking=true,before_merging=false)," "max_states=200000))"]), (455, ["--search", "let(lmc, landmark_cost_partitioning(lm_merged([lm_rhw(),lm_hm(m=1)]))," "astar(lmc,lazy_evaluator=lmc))"]), (569, ["--search", "astar(lmcut())"]), ]

这段代码定义了一个名为OPTIMAL的布尔变量，其值为True。然后，定义了一个名为CONFIGS的列表，其中包含了四个元组，每个元组包含两个元素。第一个元素是一个整数，第二个元素是一个包含命令行参数的字符串列表...

import heapq import copy # 定义状态类 class State: def init(self, board, moves=0, parent=None, last_move=None): self.board = board self.moves = moves self.parent = parent self.last_move = last_move def lt(self, other): return self.moves < other.moves def eq(self, other): return self.board == other.board # 定义转移函数 def move(state, direction): new_board = copy.deepcopy(state.board) for i in range(len(new_board)): if 0 in new_board[i]: j = new_board[i].index(0) break if direction == "up": if i == 0: return None else: new_board[i][j], new_board[i-1][j] = new_board[i-1][j], new_board[i][j] elif direction == "down": if i == len(new_board)-1: return None else: new_board[i][j], new_board[i+1][j] = new_board[i+1][j], new_board[i][j] elif direction == "left": if j == 0: return None else: new_board[i][j], new_board[i][j-1] = new_board[i][j-1], new_board[i][j] elif direction == "right": if j == len(new_board)-1: return None else: new_board[i][j], new_board[i][j+1] = new_board[i][j+1], new_board[i][j] return State(new_board, state.moves+1, state, direction) # 定义A*算法 def astar(start, goal): heap = [] closed = set() heapq.heappush(heap, start) while heap: state = heapq.heappop(heap) if state.board == goal: path = [] while state.parent: path.append(state) state = state.parent path.append(state) return path[::-1] closed.add(state) for direction in ["up", "down", "left", "right"]: child = move(state, direction) if child is None: continue if child in closed: continue if child not in heap: heapq.heappush(heap, child) else: for i, (p, c) in enumerate(heap): if c == child and p.moves > child.moves: heap[i] = (child, child) heapq.heapify(heap) # 测试 start_board = [[1, 2, 3], [4, 5, 6], [7, 8, 0]] goal_board = [[2, 3, 6], [1, 5, 8], [4, 7, 0]] start_state = State(start_board) goal_state = State(goal_board) path = astar(start_state, goal_board) for state in path: print(state.board) 这段代码运行后报错，因为State是不可hash的，那么如何修改才能使得功能一样且能够运行

import heapq 和 import copy 是 Python...因此，如果要复制一个对象并在原对象的基础上进行修改，就需要使用 copy 模块中的函数进行复制操作。copy 模块中提供了浅复制和深复制两种方式，具体的区别可以参见官方文档。

请给一下代码加注释，越详细越好。AStar.h:#ifndef ASTAR_H #define ASTAR_H #include <vector> using namespace std; class AStar { public: AStar(int n); void add_edge(int u, int v, int w); void set_heuristic(vector<int>& h); void shortest_path(int s, int t); vector<int> get_dist(); vector<int> get_prev(); private: struct edge { int to, weight; edge(int t, int w) : to(t), weight(w) {} }; int n; vector<vector<edge>> graph; vector<vector<edge>> rev_graph; vector<int> dist; vector<int> prev; vector<int> heuristic; }; class Astar { }; #endif;AStar.cpp:#include "AStar.h" #include <vector> #include <queue> #include using namespace std; AStar::AStar(int n) : n(n), graph(n), rev_graph(n), dist(n, numeric_limits<int>::max()), prev(n, -1), heuristic(n, 0) {} void AStar::add_edge(int u, int v, int w) { graph[u].push_back(edge(v, w)); rev_graph[v].push_back(edge(u, w)); } void AStar::set_heuristic(vector<int>& h) { heuristic = h; } void AStar::shortest_path(int s, int t) { priority_queue, vector>, greater>> pq; dist[s] = 0; pq.push(make_pair(heuristic[s], s)); while (!pq.empty()) { int u = pq.top().second; pq.pop(); if (u == t) return; for (auto& e : graph[u]) { int v = e.to; int w = e.weight; if (dist[v] > dist[u] + w) { dist[v] = dist[u] + w; prev[v] = u; pq.push(make_pair(dist[v] + heuristic[v], v)); } } for (auto& e : rev_graph[u]) { int v = e.to; int w = e.weight; if (dist[v] > dist[u] + w) { dist[v] = dist[u] + w; prev[v] = u; pq.push(make_pair(dist[v] + heuristic[v], v)); } } } } vector<int> AStar::get_dist() { return dist; } vector<int> AStar::get_prev() { return prev; }

注释已添加在代码中，以//开头。 AStar.h: cpp #ifndef ASTAR_H #define ASTAR_H #include using namespace std; class AStar { public: //构造函数，传入节点数n AStar(int n); //添加一条从u到v，边...

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...

def astar(initial: T, goal_test: Callable[[T], bool], successors: Callable[[T], List[T]], heuristic: Callable[[T], float]) -> Optional[Node[T]]: frontier: PriorityQueue[Node[T]] = PriorityQueue() frontier.push(Node(initial, None, 0.0, heuristic(initial))) explored: Dict[T, float] = {initial: 0.0} while not frontier.empty: current_node: Node[T] = frontier.pop() current_state: T = current_node.state if goal_test(current_state): return current_node for child in successors(current_state): new_cost: float = current_node.cost + 1 if child not in explored or explored[child] > new_cost: explored[child] = new_cost frontier.push(Node(child, current_node, new_cost, heuristic(child))) return None这段代码每行什么意思帮我加上注释

# 定义 astar 函数 def astar(initial: T, goal_test: Callable[[T], bool], successors: Callable[[T], List[T]], heuristic: Callable[[T], float]) -> Optional[Node[T]]: # 创建一个优先队列，用于存储节点 ...

Astar_八数码C语言_Astar算法_AStar_

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