結果
問題 |
No.2026 Yet Another Knapsack Problem
|
ユーザー |
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提出日時 | 2025-06-12 21:04:58 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 1,821 bytes |
コンパイル時間 | 233 ms |
コンパイル使用メモリ | 82,176 KB |
実行使用メモリ | 78,080 KB |
最終ジャッジ日時 | 2025-06-12 21:07:27 |
合計ジャッジ時間 | 51,330 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 17 WA * 17 TLE * 2 -- * 6 |
ソースコード
def main(): import sys input = sys.stdin.read().split() idx = 0 N = int(input[idx]) idx += 1 items = [] for _ in range(N): c = int(input[idx]) v = int(input[idx+1]) items.append((c, v)) idx += 2 INF = -10**18 dp = [ [INF] * (N + 1) for _ in range(N + 1) ] dp[0][0] = 0 for i in range(N): c_i, v_i = items[i] weight = i + 1 # since type is 1-based # Iterate in reverse to avoid overwriting the current state # We need to process all possible m, which can be up to c_i # But for each m, we can process it as a separate step for m in range(0, c_i + 1): # For each possible m, process the DP # We can optimize by limiting m such that m * weight <= N if m * weight > N: continue # Process in reverse order to prevent using the same m multiple times for k_prev in range(N, -1, -1): for w_prev in range(N, -1, -1): if dp[k_prev][w_prev] == INF: continue new_k = k_prev + m new_w = w_prev + m * weight if new_k > N or new_w > N: continue new_val = dp[k_prev][w_prev] + m * v_i if new_val > dp[new_k][new_w]: dp[new_k][new_w] = new_val # For each k, find the maximum value over all w <= N result = [0] * (N + 1) # result[0] unused for k in range(1, N + 1): max_val = INF for w in range(0, N + 1): if dp[k][w] > max_val: max_val = dp[k][w] result[k] = max_val for k in range(1, N + 1): print(result[k]) if __name__ == '__main__': main()