結果

問題 No.2325 Skill Tree
ユーザー googol_S0
提出日時 2023-05-28 14:12:52
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
TLE  
実行時間 -
コード長 2,490 bytes
コンパイル時間 178 ms
コンパイル使用メモリ 13,184 KB
実行使用メモリ 127,232 KB
最終ジャッジ日時 2024-12-27 00:09:07
合計ジャッジ時間 100,909 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 10 TLE * 26
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

def scc(N,edges):
M=len(edges)
start=[0]*(N+1)
elist=[0]*M
for e in edges:
start[e[0]+1]+=1
for i in range(1,N+1):
start[i]+=start[i-1]
counter=start[:]
for e in edges:
elist[counter[e[0]]]=e[1]
counter[e[0]]+=1
visited=[]
low=[0]*N
Ord=[-1]*N
ids=[0]*N
NG=[0,0]
def dfs(v):
stack=[(v,-1,0),(v,-1,1)]
while stack:
v,bef,t=stack.pop()
if t:
if bef!=-1 and Ord[v]!=-1:
low[bef]=min(low[bef],Ord[v])
stack.pop()
continue
low[v]=NG[0]
Ord[v]=NG[0]
NG[0]+=1
visited.append(v)
for i in range(start[v],start[v+1]):
to=elist[i]
if Ord[to]==-1:
stack.append((to,v,0))
stack.append((to,v,1))
else:
low[v]=min(low[v],Ord[to])
else:
if low[v]==Ord[v]:
while(True):
u=visited.pop()
Ord[u]=N
ids[u]=NG[1]
if u==v:
break
NG[1]+=1
low[bef]=min(low[bef],low[v])
for i in range(N):
if Ord[i]==-1:
dfs(i)
for i in range(N):
ids[i]=NG[1]-1-ids[i]
group_num=NG[1]
counts=[0]*group_num
for x in ids:
counts[x]+=1
groups=[[] for i in range(group_num)]
for i in range(N):
groups[ids[i]].append(i)
return groups
N=int(input())
G=[]
X=[(0,-1)]
INF=10**15
for i in range(1,N):
l,a=map(int,input().split())
X.append((l,a-1))
G.append((i,a-1))
S=scc(N,G)
for i in range(len(S)):
if len(S[i])>1:
for j in range(len(S[i])):
X[S[i][j]]=(INF,0)
import sys
sys.setrecursionlimit(10**6)
L=[-1]*N
G2=[[] for i in range(N)]
for i in range(1,N):
G2[i].append(X[i][1])
def dfs(n):
if L[n]>=0:
return L[n]
L[n]=X[n][0]
for i in range(len(G2[n])):
L[n]=max(L[n],dfs(G2[n][i]))
return L[n]
for i in range(N):
dfs(i)
C=sorted(L)
from bisect import *
Q=int(input())
for i in range(Q):
t,x=map(int,input().split())
if t==1:
print(bisect_right(C,x))
else:
if L[x-1]>=INF:
print(-1)
else:
print(L[x-1])
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