結果
| 問題 |
No.2026 Yet Another Knapsack Problem
|
| コンテスト | |
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 15:56:59 |
| 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 | 174 ms |
| コンパイル使用メモリ | 82,688 KB |
| 実行使用メモリ | 78,592 KB |
| 最終ジャッジ日時 | 2025-06-12 15:58:27 |
| 合計ジャッジ時間 | 48,564 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()
gew1fw