結果
問題 |
No.3078 Difference Sum Query
|
ユーザー |
![]() |
提出日時 | 2025-05-14 13:11:02 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,252 bytes |
コンパイル時間 | 243 ms |
コンパイル使用メモリ | 82,688 KB |
実行使用メモリ | 51,968 KB |
最終ジャッジ日時 | 2025-05-14 13:13:02 |
合計ジャッジ時間 | 5,022 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge5 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
sample | WA * 2 |
other | WA * 26 |
ソースコード
import sys def solve(): """ Solves the Very Simple Traveling Salesman Problem. Reads the number of vertices N and edges M from the input. Since the problem specifies that the graph is complete (M = N*(N-1)/2) and all edge weights are 1, any Hamiltonian cycle (visiting each vertex exactly once and returning to the start) will consist of exactly N edges. The total weight of any such cycle is the sum of the weights of its N edges. Since each edge has weight 1, the total weight is N * 1 = N. Therefore, the minimum weight of such a cycle is always N. The code only needs to read N and print it. The edge details are not required for the calculation. """ # Read N and M from the first line of input # n, m = map(int, sys.stdin.readline().split()) # We only actually need N line1 = sys.stdin.readline().split() n = int(line1[0]) # The rest of the input lines describe the edges, but we don't need them. # for _ in range(m): # sys.stdin.readline() # Read and discard edge lines if needed # The minimum weight of a Hamiltonian cycle in a complete graph # where all edge weights are 1 is simply N. print(n) if __name__ == "__main__": solve()