結果
問題 |
No.2026 Yet Another Knapsack Problem
|
ユーザー |
![]() |
提出日時 | 2025-06-12 21:10:27 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 1,745 bytes |
コンパイル時間 | 383 ms |
コンパイル使用メモリ | 81,980 KB |
実行使用メモリ | 133,672 KB |
最終ジャッジ日時 | 2025-06-12 21:12:22 |
合計ジャッジ時間 | 38,972 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 35 TLE * 1 -- * 6 |
ソースコード
def main(): import sys input = sys.stdin.read data = input().split() idx = 0 N = int(data[idx]) idx += 1 items = [] for i in range(1, N+1): c_i = int(data[idx]) v_i = int(data[idx+1]) idx += 2 items.append((i, c_i, v_i)) # Initialize DP INF = -1 << 60 dp = [[INF] * (N + 1) for _ in range(N + 1)] dp[0][0] = 0 for i, c_i, v_i in items: # We need to handle taking x items, x can be 0~c_i # Use binary optimization virtual = [] x = 1 while x <= c_i: take = min(x, c_i) virtual.append((take, take * i, take * v_i)) c_i -= take x *= 2 if c_i > 0: virtual.append((c_i, c_i * i, c_i * v_i)) # For each virtual item, process 0-1 knapsack style for dx, dw, dv in virtual: # Update dp: for each (k, w), try adding dx, dw, dv for k in range(N, -1, -1): for w in range(N, -1, -1): if dp[k][w] != INF: new_k = k + dx new_w = w + dw if new_k > N or new_w > N: continue if dp[new_k][new_w] < dp[k][w] + dv: dp[new_k][new_w] = dp[k][w] + dv # Now, for each k from 1 to N, find the maximum value among w <= N result = [] for k in range(1, N+1): max_val = INF for w in range(0, N+1): if dp[k][w] > max_val and dp[k][w] != INF: max_val = dp[k][w] result.append(max_val) for val in result: print(val) if __name__ == '__main__': main()