結果
問題 |
No.1690 Power Grid
|
ユーザー |
![]() |
提出日時 | 2025-08-10 22:31:38 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 468 ms / 3,000 ms |
コード長 | 2,785 bytes |
コンパイル時間 | 305 ms |
コンパイル使用メモリ | 82,712 KB |
実行使用メモリ | 80,436 KB |
最終ジャッジ日時 | 2025-08-10 22:31:45 |
合計ジャッジ時間 | 6,228 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 25 |
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)) from itertools import combinations N,M,K = map(int,input().split()) A = list(map(int,input().split())) XYZ = [list(map(int,input().split())) for _ in range(M)] INF = 10 ** 11 cost = [[INF] * N for _ in range(N)] for i in range(N): cost[i][i] = 0 for x,y,z in XYZ: x -= 1 y -= 1 cost[x][y] = z cost[y][x] = z for k in range(N): for i in range(N): for j in range(N): if cost[i][k]!=INF and cost[k][j]!=INF: cost[i][j] = min(cost[i][j], cost[i][k] + cost[k][j]) Q = [] for i in range(N): for j in range(i+1,N): Q.append((cost[i][j],i,j)) Q.sort() ans = INF for C in combinations(range(N),K): tmp = 0 C = set(C) for c in C: tmp += A[c] dsu = DSU(N) cnt = 0 for c,a,b in Q: if a in C and b in C: if dsu.same(a,b): continue dsu.merge(a,b) tmp += c cnt += 1 if cnt == K: break ans = min(ans,tmp) print(ans)