結果

問題 No.2536 同値性と充足可能性
ユーザー Mao-betaMao-beta
提出日時 2024-08-16 22:49:45
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 441 ms / 2,000 ms
コード長 4,785 bytes
コンパイル時間 380 ms
コンパイル使用メモリ 82,848 KB
実行使用メモリ 163,048 KB
最終ジャッジ日時 2024-08-16 22:49:54
合計ジャッジ時間 8,466 ms
ジャッジサーバーID
(参考情報)
judge2 / judge1
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ファイルパターン 結果
other AC * 31
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

import sys
import math
import bisect
from heapq import heapify, heappop, heappush
from collections import deque, defaultdict, Counter
from functools import lru_cache
from itertools import accumulate, combinations, permutations, product
sys.set_int_max_str_digits(10 ** 6)
sys.setrecursionlimit(1000000)
MOD = 10 ** 9 + 7
MOD99 = 998244353
input = lambda: sys.stdin.readline().strip()
NI = lambda: int(input())
NMI = lambda: map(int, input().split())
NLI = lambda: list(NMI())
SI = lambda: input()
SMI = lambda: input().split()
SLI = lambda: list(SMI())
EI = lambda m: [NLI() for _ in range(m)]
from typing import List, Tuple
def two_sat(n: int , clause: List[Tuple[int, bool, int, bool]]) -> List[bool] | None:
"""
clause: [(i, tf_i, j, tf_j), (), ...]
(CNF) (y1∨y2)∧(y3∨y4)∧...∧(y2m−1∨y2m)
(SCC)
#
N x1,...,xN xi ¬xi 2 2N
(a∨b) (¬a⇒b)∧(¬b⇒a) ¬a b ¬b a
v0 v1 v0 v1
#
SCC xi ¬xi
#
¬xi xi xi
# 使
ans=two_sat(N,clause)
ansNone
ansNTrueFalse
"""
answer=[0]*n
edges=[]
N=2*n
for s in clause:
i,f,j,g=s
edges.append((2*i+(0 if f else 1),2*j+(1 if g else 0)))
edges.append((2*j+(0 if g else 1),2*i+(1 if f else 0)))
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]
for i in range(n):
if ids[2*i]==ids[2*i+1]:
return None
answer[i]=(ids[2*i]<ids[2*i+1])
return answer
def main():
N, M = NMI()
clause = []
for _ in range(M):
i, e, j = SMI()
i = int(i) - 1
j = int(j) - 1
if "/" not in e:
clause.append((i, True, j, False))
clause.append((i, False, j, True))
else:
clause.append((i, True, j, True))
clause.append((i, False, j, False))
ans = two_sat(N, clause)
if ans is None:
print("No")
else:
if ans.count(True) >= (N+1)//2:
ans = [i for i, b in enumerate(ans, start=1) if b]
else:
ans = [i for i, b in enumerate(ans, start=1) if not b]
print("Yes")
print(len(ans))
print(*ans)
if __name__ == "__main__":
main()
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