题解 | #父子情深#

对于query中的节点我们记录其改变值,并通过bfs依次将其传递给下一层节点即可

# class Point:
#     def __init__(self, a=0, b=0):
#         self.x = a
#         self.y = b

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 从 1 到 n 每个结点的权值。
# @param n int 
# @param Edge Point一维数组 (u, v) 集合
# @param q int 
# @param Query Point一维数组 Point.x 为题中 r, Point.y为题中 v
# @return long一维数组
#
import collections
class Solution:
    def solve(self , n , Edge , q , Query ):
        # write code here
        weight = [0] * (n + 1)

        graph = collections.defaultdict(list)
        for i in range(len(Edge)):
            x,y = Edge[i].x,Edge[i].y
            graph[x].append(y)
            graph[y].append(x)

        change = collections.defaultdict(int)
        for i in range(len(Query)):
            x,y = Query[i].x,Query[i].y
            change[x] += y
        queue = []
        weight[1] += change[1]
        queue.append([1,weight[1]])
        visited = {1}
        while queue:
            cur,d = queue.pop(0)
            for next_ in graph[cur]:
                if next_ in visited:
                    continue
                visited.add(next_)
                weight[next_] += change[next_] + d
                queue.append([next_,weight[next_]])

        return weight[1:]