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 16:50:37 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 1,928 bytes |
| コンパイル時間 | 223 ms |
| コンパイル使用メモリ | 82,412 KB |
| 実行使用メモリ | 251,912 KB |
| 最終ジャッジ日時 | 2025-06-12 16:52:17 |
| 合計ジャッジ時間 | 33,664 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / 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().split()
idx = 0
N = int(input[idx])
idx += 1
groups = []
for i in range(1, N+1):
c_i = int(input[idx])
v_i = int(input[idx+1])
idx += 2
groups.append((i, c_i, v_i))
# Separate group 1 (i=1)
group1 = groups[0]
i1, c1, v1 = group1
assert i1 == 1 and c1 == N
other_groups = groups[1:]
# Initialize DP[x][s] = max value using x items from other groups, sum (i-1)*t_i = s
INF = -1 << 60
dp = [[INF] * (N + 1) for _ in range(N + 1)]
dp[0][0] = 0
for i, c_i, v_i in other_groups:
weight = i
value_diff = v_i - v1
excess_per = weight - 1
# Process this group
# We need to consider taking t items from this group, t from 0 to c_i
# For each possible t, update the DP
# To handle this efficiently, we can use a temporary array and iterate backwards
temp = [row[:] for row in dp]
for t in range(1, c_i + 1):
t_excess = t * excess_per
t_value = t * value_diff
for x in range(N, t - 1, -1):
for s in range(N, t_excess - 1, -1):
if temp[x - t][s - t_excess] != INF:
if dp[x][s] < temp[x - t][s - t_excess] + t_value:
dp[x][s] = temp[x - t][s - t_excess] + t_value
# Precompute the maximum for each x and s
# For each k, the answer is k*v1 + max(dp[x][s] where x <=k and s <= N -k)
for k in range(1, N + 1):
max_extra = INF
for x in range(0, k + 1):
max_s = N - k
if max_s < 0:
continue
for s in range(0, max_s + 1):
if dp[x][s] > max_extra:
max_extra = dp[x][s]
print(max_extra + k * v1)
if __name__ == '__main__':
main()
gew1fw