結果
| 問題 |
No.977 アリス仕掛けの摩天楼
|
| ユーザー |
Coki628
|
| 提出日時 | 2020-05-06 22:30:35 |
| 言語 | Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1) |
| 結果 |
AC
|
| 実行時間 | 500 ms / 2,000 ms |
| コード長 | 3,543 bytes |
| コンパイル時間 | 89 ms |
| コンパイル使用メモリ | 13,184 KB |
| 実行使用メモリ | 54,872 KB |
| 最終ジャッジ日時 | 2024-07-03 09:11:21 |
| 合計ジャッジ時間 | 4,497 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 26 |
ソースコード
import sys
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
EPS = 10 ** -10
class UnionFind:
""" Union-Find木 """
def __init__(self, n):
self.n = n
self.par = [i for i in range(n)]
self.rank = [0] * n
self.size = [1] * n
self.tree = [True] * n
self.grpcnt = n
def find(self, x):
""" 根の検索(グループ番号の取得) """
t = []
while self.par[x] != x:
t.append(x)
x = self.par[x]
for i in t:
self.par[i] = x
return self.par[x]
def union(self, x, y):
""" 併合 """
x = self.find(x)
y = self.find(y)
if x == y:
self.tree[x] = False
return
if not self.tree[x] or not self.tree[y]:
self.tree[x] = self.tree[y] = False
self.grpcnt -= 1
if self.rank[x] < self.rank[y]:
self.par[x] = y
self.size[y] += self.size[x]
else:
self.par[y] = x
self.size[x] += self.size[y]
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
def is_same(self, x, y):
""" 同じ集合に属するか判定 """
return self.find(x) == self.find(y)
def get_size(self, x=None):
if x is not None:
""" あるノードの属する集合のノード数 """
return self.size[self.find(x)]
else:
""" 集合の数 """
return self.grpcnt
def is_tree(self, x):
""" 木かどうかの判定 """
return self.tree[self.find(x)]
def SCC(N, edges):
""" 強連結成分分解 """
nodes1 = [[] for i in range(N)]
nodes2 = [[] for i in range(N)]
for u, v in edges:
nodes1[u].append(v)
nodes2[v].append(u)
T = []
visited = [False] * N
def rec1(cur):
visited[cur] = True
for nxt in nodes1[cur]:
if not visited[nxt]:
rec1(nxt)
# 行き止まったところから順にTに入れていく
T.append(cur)
# グラフが連結とは限らないので全頂点やる
for u in range(N):
if not visited[u]:
rec1(u)
visited = [False] * N
group = [0] * N
grpnum = 0
def rec2(cur):
group[cur] = grpnum
visited[cur] = True
for nxt in nodes2[cur]:
if not visited[nxt]:
rec2(nxt)
# 逆順で進めるところまで行く
for u in reversed(T):
if not visited[u]:
rec2(u)
grpnum += 1
return grpnum, group
N = INT()
edges = []
uf = UnionFind(N)
for i in range(N-1):
a, b = MAP()
edges.append((a, b))
uf.union(a, b)
if uf.get_size() == 1:
print('Bob')
elif uf.get_size() >= 3:
print('Alice')
else:
grpcnt, group = SCC(N, edges)
if grpcnt == 2:
print('Bob')
else:
print('Alice')
Coki628