結果

問題 No.2028 Even Choice
ユーザー gew1fw
提出日時 2025-06-12 16:57:57
言語 PyPy3
(7.3.15)
結果
WA  
実行時間 -
コード長 4,115 bytes
コンパイル時間 307 ms
コンパイル使用メモリ 82,460 KB
実行使用メモリ 108,464 KB
最終ジャッジ日時 2025-06-12 16:58:06
合計ジャッジ時間 7,986 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample -- * 3
other WA * 3 TLE * 1 -- * 24
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ソースコード

diff # 

import sys
import heapq

def main():
    sys.setrecursionlimit(1 << 25)
    N, K = map(int, sys.stdin.readline().split())
    A = list(map(int, sys.stdin.readline().split()))

    # We represent the array as a list of segments.
    # Each segment is a tuple: (start, end, shift, max_even, max_even_pos)
    # Initially, the array is one segment covering 0..N-1 with shift 0.
    segments = []
    heapq.heapify(segments)
    # We push the initial segment with a dummy value. The actual max_even is computed later.
    heapq.heappush(segments, (0, -1, 0, 0, 0))  # (max_even, index, shift, start, end)
    # segments is a max-heap, so we store negative values.

    # To handle the correct order, we'll process the segments and compute their max_even.

    # However, since we need to process all segments to compute their max_even, perhaps it's better to build the initial segments correctly.

    # Rebuilding the approach: Each segment is a start and end index in the original array, with a shift value indicating the number of elements removed before it. The max_even is the maximum A_i in even positions (considering the shift).

    # Let's build the initial segments correctly.

    # We'll represent each segment with a start and end (original indices), and a shift (number of elements removed before this segment). The max_even is the maximum A_i in even positions (original index - shift).

    initial_segment = (0, N-1, 0, 0, 0)  # (start, end, shift, max_even, max_even_pos)
    initial_max = 0
    initial_max_pos = -1
    for i in range(initial_segment[0], initial_segment[1]+1):
        current_pos = i - initial_segment[2]
        if current_pos % 2 == 1:  # 1-based even is 2,4,... which is 0-based index 1,3, etc.
            if A[i] > initial_max:
                initial_max = A[i]
                initial_max_pos = i
    initial_segment = (0, N-1, 0, initial_max, initial_max_pos)
    heapq.heappush(segments, (-initial_segment[3], initial_segment[0], initial_segment[1], initial_segment[2], initial_segment[4]))

    sum_total = 0
    for _ in range(K):
        if not segments:
            break  # should not happen as K <= N-1
        current = heapq.heappop(segments)
        current_max = -current[0]
        start = current[1]
        end = current[2]
        shift = current[3]
        max_pos = current[4]

        if max_pos == -1:
            # No even position in this segment, so skip
            continue
        sum_total += A[max_pos]
        # Now, split this segment into left and right
        left_start = start
        left_end = max_pos - 1
        right_start = max_pos + 1
        right_end = end

        # Update left segment
        if left_start <= left_end:
            # Compute max_even for left segment
            new_shift = shift
            max_even = 0
            max_even_pos = -1
            for i in range(left_start, left_end + 1):
                current_pos = i - new_shift
                if current_pos % 2 == 1:
                    if A[i] > max_even:
                        max_even = A[i]
                        max_even_pos = i
            # Push left segment into heap
            if max_even != 0 or max_even_pos != -1:
                heapq.heappush(segments, (-max_even, left_start, left_end, new_shift, max_even_pos))

        # Update right segment
        if right_start <= right_end:
            # Compute max_even for right segment
            new_shift = shift + 1  # because max_pos is removed, elements to the right have their positions shifted left by 1
            max_even = 0
            max_even_pos = -1
            for i in range(right_start, right_end + 1):
                current_pos = i - new_shift
                if current_pos % 2 == 1:
                    if A[i] > max_even:
                        max_even = A[i]
                        max_even_pos = i
            # Push right segment into heap
            if max_even != 0 or max_even_pos != -1:
                heapq.heappush(segments, (-max_even, right_start, right_end, new_shift, max_even_pos))

    print(sum_total)

if __name__ == '__main__':
    main()
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