Problem:
 Score : 400 points    Problem Statement Given is a rooted tree with N vertices numbered 1 to N .
The root is Vertex 1 , and the i -th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i .  Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0 .  Now, the following Q operations will be performed:   Operation j  (1 \leq j \leq Q) : Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j .   Find the value of the counter on each vertex after all operations.      Constraints  2 \leq N \leq 2 \times 10^5  1 \leq Q \leq 2 \times 10^5  1 \leq a_i < b_i \leq N  1 \leq p_j \leq N  1 \leq x_j \leq 10^4  The given graph is a tree.  All values in input are integers.         Input Input is given from Standard Input in the following format:  N  Q  a_1  b_1  :  a_{N-1}  b_{N-1}  p_1  x_1  :  p_Q  x_Q       Output Print the values of the counters on Vertex 1, 2, \ldots, N after all operations, in this order, with spaces in between.        Sample Input 1 4 3
1 2
2 3
2 4
2 10
1 100
3 1      Sample Output 1 100 110 111 110  The tree in this input is as follows:   Each operation changes the values of the counters on the vertices as follows:   Operation 1 : Increment by 10 the counter on every vertex contained in the subtree rooted at Vertex 2 , that is, Vertex 2, 3, 4 . The values of the counters on Vertex 1, 2, 3, 4 are now 0, 10, 10, 10 , respectively.  Operation 2 : Increment by 100 the counter on every vertex contained in the subtree rooted at Vertex 1 , that is, Vertex 1, 2, 3, 4 . The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 110, 110 , respectively.  Operation 3 : Increment by 1 the counter on every vertex contained in the subtree rooted at Vertex 3 , that is, Vertex 3 . The values of the counters on Vertex 1, 2, 3, 4 are now 100, 110, 111, 110 , respectively.        Sample Input 2 6 2
1 2
1 3
2 4
3 6
2 5
1 10
1 10      Sample Output 2 20 20 20 20 20 20

Solution (Python):
from collections import deque
N, Q = map(int, input().split())
G = [[] for _ in range(N)]
count = [0]*N
for i in range(N-1):
    a, b = map(int, input().split())
    G[a-1].append(b-1)
    G[b-1].append(a-1)
for i in range(Q):
    p, x = map(int, input().split())
    count[p-1]+=x
q = deque([0])
flag = [False]*N
while q:
    v = q.pop()
    flag[v] = True
    for e in G[v]:
        if flag[e]:
            continue
        count[e]+=count[v]
        q.append(e)
print(*count)

Solution (Python):
import math, string, itertools, fractions, heapq, collections, re,  array, bisect, sys, random, time, copy, functools, os, pdb
sys.setrecursionlimit(10**7)
inf = 10 ** 20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1, 0), (0, 1), (1, 0), (0, -1)]
ddn = [(-1, 0), (-1, 1), (0, 1), (1, 1), (1, 0), (1, -1), (0, -1), (-1, -1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def _I(): return int(sys.stdin.readline())
def _F(): return float(sys.stdin.readline())
def _pf(s): return print(s, flush=True)
def perr(s):
    if 'LOCAL' in os.environ:
        print(s)

def main():
    global PX, nodes
    if 'BIG' in os.environ:
        N = 2* 10** 5
        Q = 2 * 10 ** 5
        AB = [[i, i + 1] for i in range(1, N)]
        PX = [[i, 10] for i in range(Q)]
    else:
        N, Q = LI()
        
        nodes = [[] for _ in range(N+1)]
        PX = [0] * (N+1)
        for i in range(N-1):
            a, b = LI()
            
            nodes[a].append(b)
            nodes[b].append(a)

        for i in range(Q):
            p, x = LI()
            PX[p] += x
        dfs(1, -1)
        print(' '.join(map(lambda x: str(x), PX[1:])))
        
        
        

def dfs(parent, parent_parent):
    for node in nodes[parent]:
        if node != parent_parent:
            PX[node] += PX[parent]
            dfs(node, parent)

if __name__ == '__main__':
    main()