結果

問題 No.479 頂点は要らない
ユーザー lam6er
提出日時 2025-03-26 15:52:34
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 1,809 bytes
コンパイル時間 335 ms
コンパイル使用メモリ 82,192 KB
実行使用メモリ 111,588 KB
最終ジャッジ日時 2025-03-26 15:53:04
合計ジャッジ時間 5,791 ms
ジャッジサーバーID
(参考情報)
judge4 / 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 WA * 3
other WA * 38
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 main():
import sys
input = sys.stdin.read().split()
idx = 0
n = int(input[idx])
idx += 1
m = int(input[idx])
idx += 1
edges = [[] for _ in range(n)]
for _ in range(m):
a = int(input[idx])
idx += 1
b = int(input[idx])
idx += 1
if a > b:
a, b = b, a
edges[a].append(b)
edges[b].append(a)
selected = [False] * n
# Initialize selected with all the u of each edge (u, v)
for u in range(n):
for v in edges[u]:
if u < v:
selected[u] = True
# Iterate vertices from largest to smallest
for i in reversed(range(n)):
if not selected[i]:
continue
can_remove = True
# Check all adjacent edges
for j in edges[i]:
if j < i:
# This edge is stored in the j's list as u, but here i is the larger one
# The edge is (j, i), so if j is not selected, then i must be kept
if not selected[j]:
can_remove = False
break
else:
# The edge is (i, j), where i is the smaller one. But since we are processing i in reverse order, j could be larger.
# Since i was initially selected as the smaller vertex, but if we remove i, we need to check if j is selected.
if not selected[j]:
can_remove = False
break
if can_remove:
selected[i] = False
# Calculate the sum
total = 0
for i in range(n):
if selected[i]:
total += (1 << i)
# Convert to binary
if total == 0:
print("0")
else:
print(bin(total)[2:])
print()
if __name__ == '__main__':
main()
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0