結果
| 問題 |
No.748 yuki国のお財布事情
|
| コンテスト | |
| ユーザー |
ntuda
|
| 提出日時 | 2025-06-23 21:37:00 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
AC
|
| 実行時間 | 580 ms / 2,000 ms |
| コード長 | 2,234 bytes |
| コンパイル時間 | 335 ms |
| コンパイル使用メモリ | 82,172 KB |
| 実行使用メモリ | 104,188 KB |
| 最終ジャッジ日時 | 2025-06-23 21:37:09 |
| 合計ジャッジ時間 | 9,065 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 |
|---|---|
| sample | AC * 3 |
| other | AC * 26 |
ソースコード
import typing
class DSU:
'''
Implement (union by size) + (path halving)
Reference:
Zvi Galil and Giuseppe F. Italiano,
Data structures and algorithms for disjoint set union problems
'''
def __init__(self, n: int = 0) -> None:
self._n = n
self.parent_or_size = [-1] * n
def merge(self, a: int, b: int) -> int:
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[x] < -self.parent_or_size[y]:
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: int, b: int) -> bool:
assert 0 <= a < self._n
assert 0 <= b < self._n
return self.leader(a) == self.leader(b)
def leader(self, a: int) -> int:
assert 0 <= a < self._n
parent = self.parent_or_size[a]
while parent >= 0:
if self.parent_or_size[parent] < 0:
return parent
self.parent_or_size[a], a, parent = (
self.parent_or_size[parent],
self.parent_or_size[parent],
self.parent_or_size[self.parent_or_size[parent]]
)
return a
def size(self, a: int) -> int:
assert 0 <= a < self._n
return -self.parent_or_size[self.leader(a)]
def groups(self) -> typing.List[typing.List[int]]:
leader_buf = [self.leader(i) for i in range(self._n)]
result: typing.List[typing.List[int]] = [[] for _ in range(self._n)]
for i in range(self._n):
result[leader_buf[i]].append(i)
return list(filter(lambda r: r, result))
N,M,K = map(int,input().split())
dsu = DSU(N)
ABC = []
for i in range(M):
a,b,c = map(int,input().split())
a -= 1
b -= 1
ABC.append((a,b,c,i))
E = set()
for i in range(K):
e = int(input()) - 1
E.add(e)
for e in E:
a,b,_,_ = ABC[e]
dsu.merge(a,b)
ABC.sort(key = lambda x:x[2])
ans = 0
for a,b,c,i in ABC:
if i in E:
continue
if dsu.same(a,b):
ans += c
else:
dsu.merge(a,b)
print(ans)
ntuda