結果
問題 | No.1690 Power Grid |
ユーザー | zkou |
提出日時 | 2021-09-24 23:21:19 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 1,870 ms / 3,000 ms |
コード長 | 5,081 bytes |
コンパイル時間 | 550 ms |
コンパイル使用メモリ | 87,164 KB |
実行使用メモリ | 80,292 KB |
最終ジャッジ日時 | 2023-09-18 22:23:23 |
合計ジャッジ時間 | 15,945 ms |
ジャッジサーバーID (参考情報) |
judge15 / judge14 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 73 ms
71,444 KB |
testcase_01 | AC | 74 ms
71,240 KB |
testcase_02 | AC | 73 ms
71,472 KB |
testcase_03 | AC | 73 ms
71,224 KB |
testcase_04 | AC | 73 ms
71,232 KB |
testcase_05 | AC | 72 ms
71,436 KB |
testcase_06 | AC | 965 ms
79,564 KB |
testcase_07 | AC | 994 ms
79,584 KB |
testcase_08 | AC | 933 ms
79,604 KB |
testcase_09 | AC | 981 ms
79,580 KB |
testcase_10 | AC | 90 ms
76,664 KB |
testcase_11 | AC | 107 ms
77,652 KB |
testcase_12 | AC | 73 ms
71,452 KB |
testcase_13 | AC | 112 ms
77,684 KB |
testcase_14 | AC | 339 ms
78,388 KB |
testcase_15 | AC | 88 ms
76,576 KB |
testcase_16 | AC | 1,870 ms
80,132 KB |
testcase_17 | AC | 970 ms
79,720 KB |
testcase_18 | AC | 463 ms
78,244 KB |
testcase_19 | AC | 1,803 ms
79,596 KB |
testcase_20 | AC | 1,823 ms
80,292 KB |
testcase_21 | AC | 84 ms
76,508 KB |
testcase_22 | AC | 862 ms
79,552 KB |
testcase_23 | AC | 86 ms
76,508 KB |
testcase_24 | AC | 264 ms
78,648 KB |
testcase_25 | AC | 159 ms
78,132 KB |
testcase_26 | AC | 113 ms
77,576 KB |
testcase_27 | AC | 74 ms
71,412 KB |
ソースコード
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] N, M, K = map(int, input().split()) As = list(map(int, input().split())) XYZs = [tuple(map(int, input().split())) for _ in range(M)] INF = 10 ** 18 adj = [[INF] * N for _ in range(N)] for X, Y, Z in XYZs: X -= 1; Y -= 1 adj[X][Y] = adj[Y][X] = min(adj[X][Y], Z) for k in range(N): for i in range(N): for j in range(N): adj[i][j] = min(adj[i][j], adj[i][k] + adj[k][j]) def calc(indices): t = [] for idx_i, i in enumerate(indices): for idx_j, j in enumerate(indices): if idx_i == idx_j: continue t.append((adj[i][j], idx_i, idx_j)) t.sort(key=lambda t: t[0]) uf = dsu(K) ret = 0 for c, i, j in t: if not uf.same(i, j): uf.merge(i, j) ret += c return ret def popcnt(n): "https://atcoder.jp/contests/abc152/submissions/9619555" c = (n & 0x5555555555555555) + ((n >> 1) & 0x5555555555555555) c = (c & 0x3333333333333333) + ((c >> 2) & 0x3333333333333333) c = (c & 0x0f0f0f0f0f0f0f0f) + ((c >> 4) & 0x0f0f0f0f0f0f0f0f) c = (c & 0x00ff00ff00ff00ff) + ((c >> 8) & 0x00ff00ff00ff00ff) c = (c & 0x0000ffff0000ffff) + ((c >> 16) & 0x0000ffff0000ffff) c = (c & 0x00000000ffffffff) + ((c >> 32) & 0x00000000ffffffff) return c answer = INF for s in range(1 << N): if popcnt(s) != K: continue indices = [] tmp = 0 for i in range(N): if s & (1 << i): indices.append(i) tmp += As[i] tmp += calc(indices) answer = min(answer, tmp) print(answer)