結果

問題 No.2337 Equidistant
ユーザー とりゐ
提出日時 2023-06-02 22:09:23
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 2,944 ms / 4,000 ms
コード長 4,018 bytes
コンパイル時間 629 ms
コンパイル使用メモリ 82,432 KB
実行使用メモリ 242,316 KB
最終ジャッジ日時 2024-12-28 18:24:09
合計ジャッジ時間 35,888 ms
ジャッジサーバーID
(参考情報)
judge1 / judge2
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

class JumpOnTree:
def __init__(self, edges, root=0):
self.n = len(edges)
self.edges = edges
self.root = root
self.logn = (self.n - 1).bit_length()
self.depth = [-1] * self.n
self.depth[self.root] = 0
self.parent = [[-1] * self.n for _ in range(self.logn)]
self.dfs()
self.doubling()
def dfs(self):
stack = [self.root]
while stack:
u = stack.pop()
for v in self.edges[u]:
if self.depth[v] == -1:
self.depth[v] = self.depth[u] + 1
self.parent[0][v] = u
stack.append(v)
def doubling(self):
for i in range(1, self.logn):
for u in range(self.n):
p = self.parent[i - 1][u]
if p != -1:
self.parent[i][u] = self.parent[i - 1][p]
def lca(self, u, v):
du = self.depth[u]
dv = self.depth[v]
if du > dv:
du, dv = dv, du
u, v = v, u
d = dv - du
i = 0
while d > 0:
if d & 1:
v = self.parent[i][v]
d >>= 1
i += 1
if u == v:
return u
logn = (du - 1).bit_length()
for i in range(logn - 1, -1, -1):
pu = self.parent[i][u]
pv = self.parent[i][v]
if pu != pv:
u = pu
v = pv
return self.parent[0][u]
def dist(self,u,v):
L=self.lca(u,v)
return self.depth[u]+self.depth[v]-self.depth[L]*2
def jump(self, u, v, k):
if k == 0:
return u
p = self.lca(u, v)
d1 = self.depth[u] - self.depth[p]
d2 = self.depth[v] - self.depth[p]
if d1 + d2 < k:
return -1
if k <= d1:
d = k
else:
u = v
d = d1 + d2 - k
i = 0
while d > 0:
if d & 1:
u = self.parent[i][u]
d >>= 1
i += 1
return u
from sys import stdin
input=lambda :stdin.readline()[:-1]
n,q=map(int,input().split())
edge=[[] for i in range(n)]
for i in range(n-1):
a,b=map(lambda x:int(x)-1,input().split())
edge[a].append(b)
edge[b].append(a)
JT=JumpOnTree(edge)
query=[[] for i in range(n)]
for i in range(q):
s,t=map(lambda x:int(x)-1,input().split())
d=JT.dist(s,t)
if d%2:
continue
mid=JT.jump(s,t,d//2)
ng1=JT.jump(mid,s,1)
ng2=JT.jump(mid,t,1)
query[mid].append((i,ng1,ng2))
def rerooting(query):
dp=[[E]*len(edge[v]) for v in range(n)]
# dfs1
memo=[E]*n
for v in order[::-1]:
res=E
for i in range(len(edge[v])):
if edge[v][i]==par[v]:
continue
dp[v][i]=memo[edge[v][i]]
res=merge(res,f(dp[v][i],edge[v][i]))
memo[v]=g(res,v)
# dfs2
memo2=[E]*n
for v in order:
for i in range(len(edge[v])):
if edge[v][i]==par[v]:
dp[v][i]=memo2[v]
s=len(edge[v])
cumR=[E]*(s+1)
cumR[s]=E
for i in range(s,0,-1):
cumR[i-1]=merge(cumR[i],f(dp[v][i-1],edge[v][i-1]))
cumL=E
for i in range(s):
if edge[v][i]!=par[v]:
val=merge(cumL,cumR[i+1])
memo2[edge[v][i]]=g(val,v)
cumL=merge(cumL,f(dp[v][i],edge[v][i]))
ans=[0]*q
for v in range(n):
if query[v]:
dic={}
for i in range(len(edge[v])):
dic[edge[v][i]]=dp[v][i]
for i,ng1,ng2 in query[v]:
res=n
res-=dic[ng1]
res-=dic[ng2]
ans[i]=res
return ans
E=0
def f(res,v):
return res
def g(res,v):
return res+1
def merge(a,b):
return a+b
def calc_ans(res,v):
return g(res,v)
# make order table
# root = 0
from collections import deque
order=[]
par=[-1]*n
todo=deque([0])
while todo:
v=todo.popleft()
order.append(v)
for u in edge[v]:
if u!=par[v]:
par[u]=v
todo.append(u)
ans=rerooting(query)
print(*ans,sep='\n')
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0