結果

問題 No.1914 Directed by Two Sequences
ユーザー chineristAC
提出日時 2022-02-06 23:11:19
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,600 bytes
コンパイル時間 438 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 114,220 KB
最終ジャッジ日時 2024-12-16 01:35:51
合計ジャッジ時間 15,677 ms
ジャッジサーバーID
(参考情報)
judge1 / judge5
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 RE * 28
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ソースコード

diff #
プレゼンテーションモードにする

def solve(N,A,B,M,E):
A = [a-1 for a in A]
B = [b-1 for b in B]
res = set()
edge = [set() for v in range(N)]
for u,v in E:
edge[u-1].add(v-1)
edge[v-1].add(u-1)
group = [-1 for v in range(N)]
for v in range(N):
mex = [False for i in range(len(edge[v])+1)]
for nv in edge[v]:
if nv < v and group[nv] < len(mex):
mex[group[nv]] = True
for i in range(len(mex)):
if not mex[i]:
group[v] = i
break
n = max(group) + 1
clique = [[] for g in range(n)]
for v in range(N):
clique[group[v]].append(v)
def direct(i,j):
if i < j:
return A[i] < B[j]
else:
return B[i] < A[j]
def hamilton_path(V):
n = len(V)
if n <= 1:
return V
A = hamilton_path(V[:n//2])
B = hamilton_path(V[n//2:])
res = []
bi = 0
for ai in range(len(A)):
while bi<len(B) and direct(B[bi],A[ai]):
res.append(B[bi])
bi += 1
res.append(A[ai])
res += B[bi:]
return res
idx = [-1 for v in range(N)]
for g in range(n):
clique[g] = hamilton_path(clique[g])
for i in range(len(clique[g])):
idx[clique[g][i]] = i
for u,v in zip(clique[g],clique[g][1:]):
res.add((u,v))
memo_to = [N for v in range(N)]
memo_from = [-1 for v in range(N)]
ci = [i for i in range(n)]
ci.sort(lambda g:len(clique[g]))
Vs = []
for i in range(n):
target = ci[i]
Vs += clique[target]
for v in Vs:
for nv in clique[target]:
if v==nv:
continue
if direct(v,nv) and nv not in edge[v]:
memo_to[v] = min(memo_to[v],idx[nv])
if direct(nv,v) and v not in edge[nv]:
memo_from[v] = max(memo_from[v],idx[nv])
if memo_to[v]!=N:
nv = clique[target][memo_to[v]]
res.add((v,nv))
memo_to[v] = N
if memo_from[v]!=-1:
pv = clique[target][memo_from[v]]
res.add((pv,v))
memo_from[v] = -1
print(len(res))
for u,v in res:
print(u+1,v+1)
import sys
input = lambda :sys.stdin.buffer.readline()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
N,M = mi()
assert N<= 5000
A = li()
B = li()
E = [tuple(mi()) for i in range(M)]
solve(N,A,B,M,E)
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