結果
問題 | No.3078 Very Simple Traveling Salesman Problem |
ユーザー | aram14 |
提出日時 | 2021-04-01 20:30:20 |
言語 | Python3 (3.12.2 + numpy 1.26.4 + scipy 1.12.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 1,649 bytes |
コンパイル時間 | 114 ms |
コンパイル使用メモリ | 12,800 KB |
実行使用メモリ | 21,404 KB |
最終ジャッジ日時 | 2024-05-10 06:31:40 |
合計ジャッジ時間 | 3,810 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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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 | 32 ms
16,384 KB |
testcase_01 | AC | 31 ms
10,624 KB |
testcase_02 | AC | 32 ms
10,880 KB |
testcase_03 | TLE | - |
testcase_04 | -- | - |
testcase_05 | -- | - |
testcase_06 | -- | - |
testcase_07 | -- | - |
testcase_08 | -- | - |
testcase_09 | -- | - |
testcase_10 | -- | - |
ソースコード
from collections import defaultdict, namedtuple def connect(graph, u, v, w): graph[u][v] = w graph[v][u] = w def tsp(d): """solve the traveling salesman problem by bitDP Parameters ----------- d : list of list or numpy.array distance matrix; d[i][j] is the distance between i and j Returns ------- the minimum cost of TSP """ n = len(d) # number of cities # DP[A] = {v: value} # A is the set of visited cities other than v # v is the last visited city # value is the cost DP = defaultdict(dict) for A in range(1, 1 << n): if A & 1 << 0 == 0: # 0 is not included in set A continue # main for v in range(n): if A & (1 << v) == 0: # v is not included in set A if A == (1 << 0): # v is the first visited city DP[A][v] = d[0][v] if d[0][v] > 0 else float('inf') else: # A ^ (1 << u) <=> A - {u} DP[A][v] = float('inf') for u in range(1, n): if d[u][v] > 0 and A & (1 << u) != 0: DP[A][v] = min(DP[A][v], DP[A ^ (1 << u)][u] + d[u][v]) V = 1 << n DP[V][0] = float('inf') for u in range(1, n): if d[u][0] > 0 and A & (1 << u) != 0: DP[V][0] = min(DP[V][0], DP[A ^ (1 <<u)][u] + d[u][0]) return DP[V][0] # 入力 N, M = map(int, input().split()) graph = [[0 for i in range(N)] for j in range(N)] for _ in range(M): u, v, w = map(int, input().split()) connect(graph, u-1, v-1, w) res = tsp(graph) print(res)