結果
| 問題 |
No.1324 Approximate the Matrix
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 20:51:38 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,318 bytes |
| コンパイル時間 | 180 ms |
| コンパイル使用メモリ | 82,424 KB |
| 実行使用メモリ | 86,096 KB |
| 最終ジャッジ日時 | 2025-03-20 20:51:48 |
| 合計ジャッジ時間 | 6,027 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 9 WA * 33 |
ソースコード
import heapq
def main():
import sys
input = sys.stdin.read().split()
idx = 0
N, K = int(input[idx]), int(input[idx+1])
idx += 2
A = list(map(int, input[idx:idx+N]))
idx += N
B = list(map(int, input[idx:idx+N]))
idx += N
P = []
for _ in range(N):
row = list(map(int, input[idx:idx+N]))
idx += N
P.append(row)
sum_p_sq = sum(p * p for row in P for p in row)
remaining_row = A.copy()
remaining_col = B.copy()
heap = []
for i in range(N):
for j in range(N):
initial_cost = 1 - 2 * P[i][j]
heapq.heappush(heap, (initial_cost, i, j))
total_cost = 0
x = [[0] * N for _ in range(N)]
for _ in range(K):
while True:
if not heap:
break # should not happen since G is non-empty
cost, i, j = heapq.heappop(heap)
if remaining_row[i] > 0 and remaining_col[j] > 0:
x[i][j] += 1
remaining_row[i] -= 1
remaining_col[j] -= 1
total_cost += cost
new_cost = 2 * x[i][j] + 1 - 2 * P[i][j]
heapq.heappush(heap, (new_cost, i, j))
break
print(total_cost + sum_p_sq)
if __name__ == '__main__':
main()
lam6er