結果

問題 No.2991 Hypercubic Graph Flow
ユーザー chineristAC
提出日時 2024-12-16 00:26:07
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 156 ms / 2,000 ms
コード長 3,247 bytes
コンパイル時間 179 ms
コンパイル使用メモリ 82,240 KB
実行使用メモリ 88,112 KB
最終ジャッジ日時 2024-12-16 00:26:09
合計ジャッジ時間 2,048 ms
ジャッジサーバーID
(参考情報)
judge1 / judge3
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ファイルパターン 結果
other AC * 10
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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

import sys,time
from itertools import permutations
from heapq import heappop,heappush
from collections import deque
import random
import bisect
from math import log,gcd
input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())
def find_euler_tour(N,M,edge):
done_counter = [0] * N
used = [False] * M
visit = [False] * N
res = []
for start in range(N):
if visit[start]:
continue
visit[start] = True
st = [start]
tmp_res = []
while st:
v = st[-1]
visit[v] = True
if done_counter[v] == len(edge[v]):
tmp_res.append(v)
st.pop()
continue
nv,eid = edge[v][done_counter[v]]
done_counter[v] += 1
if used[eid]:
continue
used[eid] = True
st.append(nv)
res.append(tmp_res)
return res
def solve(N):
if N == 1:
return ("No",[])
if N & 1 == 0:
edge = [[] for v in range(1<<N)]
nxt_eid = 0
for v in range(1<<N):
for k in range(N):
if v>>k & 1 == 0:
edge[v].append((v^(1<<k),nxt_eid))
edge[v^(1<<k)].append((v,nxt_eid))
nxt_eid += 1
euler_tour = find_euler_tour(1<<N,nxt_eid,edge)
res = [[0]*(1<<N) for i in range(1<<N)]
for ce in euler_tour:
for u,v in zip(ce,ce[1:]):
res[u][v] = 1
res[v][u] = -1
return ("Yes",res)
else:
edge = [[] for v in range(1<<N)]
nxt_eid = 0
for v in range(1<<N):
for k in range(3,N):
if v>>k & 1 == 0:
edge[v].append((v^(1<<k),nxt_eid))
edge[v^(1<<k)].append((v,nxt_eid))
nxt_eid += 1
euler_tour = find_euler_tour(1<<N,nxt_eid,edge)
res = [[0]*(1<<N) for i in range(1<<N)]
for ce in euler_tour:
for u,v in zip(ce,ce[1:]):
res[u][v] = 1
res[v][u] = -1
for root in range(0,1<<N,8):
two_in = [0,3,5,6]
two_out = [1,2,4,7]
for a,b in zip(two_in,two_out):
res[a^root][b^root] = -2
res[b^root][a^root] = 2
for a in two_in:
for k in [2,4]:
res[a^root][a^k^root] = 1
res[a^k^root][a^root] = -1
return ("Yes",res)
def checker(N,res):
for i in range(1<<N):
for k in range(N):
if res[i][i^(1<<k)] == 0:
print(i,i^(1<<k))
return False
for j in range(1<<N):
d = i^j
if (d & (d-1)) != 0 and res[i][j]!=0:
return (False,"must 0")
if sum(res[i])!=0:
return (False,"sum")
return True
N = int(input())
ans,res = solve(N)
if ans == "Yes":
print(ans)
for i in range(1<<N):
print(*res[i])
else:
print(ans)
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