python实现cbs算法

CBS算法是一种基于冲突搜索的路径规划算法，用于解决多智能体系统的路径规划问题。在Python中实现CBS算法，可以依赖于第三方库例如networkx，numpy等。以下是基本的伪代码实现：

//定义状态类，每个状态有当前路径集合和当前冲突集合
class State:
    def __init__(self, paths, conflicts):
        self.paths = paths
        self.conflicts = conflicts

//定义CBS算法的主函数
def cbs(start, goal, constraints):
    //初始化起始状态，并将其加入open列表
    initial_state = State(path, [])
    open_list = [initial_state]
    //循环直至找到解或者open列表为空
    while open_list:
        //从open列表中弹出最小 F 值的状态
        current = min(open_list, key=lambda x: x.F)
        open_list.remove(current)
        //判断当前状态是否是目标状态
        if is_goal(current):
            return current.paths
        //扩展状态并将生成的新状态加入open列表
        for state in expand(current, constraints):
            open_list.append(state)
    return None

//定义状态扩展函数
def expand(state, constraints):
    //从当前状态获取所有的冲突
    conflicts = state.conflicts
    //循环扩展路径并查看是否生成新的冲突
    for i in range(len(state.paths)):
        for j in range(i+1, len(state.paths)):
            if conflict(state.paths[i], state.paths[j], constraints):
                //如果存在冲突就生成新路径集合和新冲突集合
                new_paths = resolve_conflict(state.paths[i], state.paths[j], conflicts)
                new_conflicts = conflicts.append((i, j, t))
                //返回新状态
                yield State(new_paths, new_conflicts)

//定义相邻路径冲突检测函数
def conflict(path1, path2, constraints):
    //定义路径1和路径2之间的交集
    intersect = set(path1) & set(path2)
    //循环检查每个时间步骤是否符合约束条件
    for t in intersect:
        for c in constraints:
            if c.check(path1, path2, t):
                //如果有任一时间步骤违反了约束条件，则判断为冲突
                return True
    //如果没有冲突返回False
    return False

//定义解决冲突函数
def resolve_conflict(path1, path2, conflicts):
    //找到所有冲突涉及到的时间步骤
    timesteps = set()
    for c in conflicts:
        if c[0] == path1 or c[1] == path1 or c[0] == path2 or c[1] == path2:
            timesteps.add(c[2])
    //将所有与路径1和路径2相关的时间步骤合并成一个新路径
    new_path = []
    for t in range(max([max(path1), max(path2)], default=0)):
        if t in timesteps:
            new_path.append(path2[t])
        else:
            new_path.append(path1[t])
    //返回合并后的新路径
    return new_path