結果
| 問題 |
No.1947 質より種類数
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-04-15 23:41:10 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,806 bytes |
| コンパイル時間 | 542 ms |
| コンパイル使用メモリ | 81,796 KB |
| 実行使用メモリ | 134,308 KB |
| 最終ジャッジ日時 | 2025-04-15 23:43:02 |
| 合計ジャッジ時間 | 8,585 ms |
|
ジャッジサーバーID (参考情報) |
judge2 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 12 WA * 2 TLE * 1 -- * 22 |
ソースコード
def main():
import sys
input = sys.stdin.read().split()
idx = 0
N = int(input[idx]); idx +=1
V = int(input[idx]); idx +=1
C = int(input[idx]); idx +=1
types = []
for _ in range(N):
v = int(input[idx]); idx +=1
w = int(input[idx]); idx +=1
types.append((v, w))
# Sort types in descending order of w/v ratio
types.sort(key=lambda x: -(x[1] / x[0]))
# Initialize DP: dp[k] is a dictionary {sum_v: (sum_w, max_v, max_w)}
dp = [{} for _ in range(N+1)]
dp[0][0] = (0, 0, 0) # 0 types, sum_v=0, sum_w=0, max_v=0, max_w=0
for i in range(N):
v_i, w_i = types[i]
# Iterate in reverse to avoid processing the same item multiple times
for k in range(N, 0, -1):
current_dp = dp[k-1]
for sum_v in list(current_dp.keys()):
sum_w, current_max_v, current_max_w = current_dp[sum_v]
new_sum_v = sum_v + v_i
if new_sum_v > V:
continue
new_sum_w = sum_w + w_i
# Determine the new max ratio
current_ratio = current_max_w / current_max_v if current_max_v != 0 else 0
new_type_ratio = w_i / v_i
if new_type_ratio > current_ratio:
new_max_v, new_max_w = v_i, w_i
else:
new_max_v, new_max_w = current_max_v, current_max_w
# Update the new state
if new_sum_v not in dp[k] or new_sum_w > dp[k].get(new_sum_v, (0, 0, 0))[0] or \
(new_sum_w == dp[k].get(new_sum_v, (0, 0, 0))[0] and (new_max_w / new_max_v if new_max_v !=0 else 0) > (dp[k][new_sum_v][2] / dp[k][new_sum_v][1] if dp[k][new_sum_v][1] !=0 else 0)):
dp[k][new_sum_v] = (new_sum_w, new_max_v, new_max_w)
# Handle the case where this is the first type added (k=1)
if v_i <= V:
new_sum_v = v_i
new_sum_w = w_i
new_k = 1
new_max_v, new_max_w = v_i, w_i
if new_sum_v not in dp[new_k] or new_sum_w > dp[new_k].get(new_sum_v, (0, 0, 0))[0]:
dp[new_k][new_sum_v] = (new_sum_w, new_max_v, new_max_w)
max_satisfaction = 0
for k in range(N+1):
for sum_v in dp[k]:
sum_w, max_v, max_w = dp[k][sum_v]
if sum_v > V:
continue
remaining = V - sum_v
if max_v == 0:
additional = 0
else:
additional = (remaining // max_v) * max_w
satisfaction = k * C + sum_w + additional
if satisfaction > max_satisfaction:
max_satisfaction = satisfaction
print(max_satisfaction)
if __name__ == "__main__":
main()
lam6er