树上战争 Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 799 Accepted Submission(s): 436 Problem Description 给一棵树,如果树上的某个节点被某个人占据,则它的所有儿子都被占据,lxh和pfz初始时分别站在两个节点上,谁当前所在的点被另一个人占据,他就输了比赛,问谁能获胜 In
将多个属性的内容更新到节点上 def update_by_id(id,graph,**kwargs): """ 更新节点的属性 根据节点的ID来更新节点的属性,如果存在该属性,则更新,如果不存在该属性,则添加 """ if graph is None: graph = get_graph() if kwargs is None: return None match = "match (x) where id(x)=%s " %