結果
| 問題 |
No.1719 Tree and Permutation
|
| コンテスト | |
| ユーザー |
 zkou
|
| 提出日時 | 2021-09-23 17:17:52 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 3,986 bytes |
| コンパイル時間 | 295 ms |
| コンパイル使用メモリ | 82,280 KB |
| 実行使用メモリ | 109,188 KB |
| 最終ジャッジ日時 | 2024-09-23 03:49:10 |
| 合計ジャッジ時間 | 3,508 ms |
|
ジャッジサーバーID （参考情報） |
judge3 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | WA * 1 |
| other | WA * 6 |
ソースコード
import sys
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
class dsu:
"""Data structures and algorithms for disjoint set union problems.
Given an undirected graph, it processes the following queries in O(alpha(n)) time (amortized).
> Edge addition
> Deciding whether given two vertices are in the same connected component
Each connected component internally has a representative vertex.
When two connected components are merged by edge addition,
one of the two representatives of these connected components becomes the representative of the new connected component.
"""
__slots__ = ["n", "parent_or_size"]
def __init__(self, n):
"""It creates an undirected graph with n vertices and 0 edges.
Constraints
-----------
> 0 <= n <= 10 ** 8
Complexity
----------
> O(n)
"""
self.n = n
self.parent_or_size = [-1] * n
def merge(self, a, b):
"""It adds an edge (a, b).
If the vertices a and b were in the same connected component,
it returns the representative of this connected component.
Otherwise, it returns the representative of the new connected component.
Constraints
-----------
> 0 <= a < n
> 0 <= b < n
Complexity
----------
> O(alpha(n)) amortized
"""
# assert 0 <= a < self.n
# assert 0 <= b < self.n
x = self.leader(a)
y = self.leader(b)
if x == y:
return x
if self.parent_or_size[y] < self.parent_or_size[x]:
x, y = y, x
self.parent_or_size[x] += self.parent_or_size[y]
self.parent_or_size[y] = x
return x
def same(self, a, b):
"""It returns whether the vertices a and b are in the same connected component.
Constraints
-----------
> 0 <= a < n
> 0 <= b < n
Complexity
----------
> O(alpha(n)) amortized
"""
# assert 0 <= a < self.n
# assert 0 <= b < self.n
return self.leader(a) == self.leader(b)
def leader(self, a):
"""It returns the representative of the connected component that contains the vertex a.
Constraints
-----------
> 0 <= a < n
Complexity
----------
> O(alpha(n)) amortized
"""
# assert 0 <= a < self.n
path = []
while self.parent_or_size[a] >= 0:
path.append(a)
a = self.parent_or_size[a]
for child in path:
self.parent_or_size[child] = a
return a
def size(self, a):
"""It returns the size of the connected component that contains the vertex a.
Constraints
-----------
> 0 <= a < n
Complexity
----------
> O(alpha(n)) amortized
"""
# assert 0 <= a < self.n
return -self.parent_or_size[self.leader(a)]
def groups(self):
"""It divides the graph into connected components and returns the list of them.
More precisely, it returns the list of the "list of the vertices in a connected component".
Both of the orders of the connected components and the vertices are undefined.
Complexity
----------
> O(n)
"""
result = [[] for _ in range(self.n)]
for i in range(self.n):
result[self.leader(i)].append(i)
return [g for g in result if g]
T = int(input())
assert 1 <= T <= 10 ** 5
N_sum = 0
for _ in range(T):
N = int(input())
N_sum += N
assert 3 <= N <= 10 ** 5
uf = dsu(N)
for _ in range(N - 1):
u, v = map(int, input().split())
u -= 1; v -= 1
assert u != v
assert 0 <= u < N
assert 0 <= v < N
uf.merge(u, v)
assert len(uf.groups()) == 1
assert N_sum <= 5 * 10 ** 5