結果
| 問題 | 
                            No.1117 数列分割
                             | 
                    
| コンテスト | |
| ユーザー | 
                             lam6er
                         | 
                    
| 提出日時 | 2025-04-16 15:59:40 | 
| 言語 | PyPy3  (7.3.15)  | 
                    
| 結果 | 
                             
                                AC
                                 
                             
                            
                         | 
                    
| 実行時間 | 2,265 ms / 3,000 ms | 
| コード長 | 2,220 bytes | 
| コンパイル時間 | 467 ms | 
| コンパイル使用メモリ | 81,856 KB | 
| 実行使用メモリ | 94,104 KB | 
| 最終ジャッジ日時 | 2025-04-16 16:03:25 | 
| 合計ジャッジ時間 | 25,724 ms | 
| 
                            ジャッジサーバーID (参考情報)  | 
                        judge5 / judge4 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 3 | 
| other | AC * 26 | 
ソースコード
import sys
from collections import deque
def main():
    import sys
    input = sys.stdin.read().split()
    idx = 0
    N = int(input[idx]); idx +=1
    K = int(input[idx]); idx +=1
    M = int(input[idx]); idx +=1
    A = list(map(int, input[idx:idx+N]))
    idx += N
    # Prefix sum array (1-based)
    S = [0] * (N + 1)
    for i in range(1, N+1):
        S[i] = S[i-1] + A[i-1]
    # Initialize DP with two arrays for space optimization
    prev_dp = [-float('inf')] * (N + 1)
    prev_dp[0] = 0  # base case: 0 elements, 0 segments
    for k in range(1, K+1):
        curr_dp = [-float('inf')] * (N + 1)
        que1 = deque()  # for max (dp[j] - S[j])
        que2 = deque()  # for max (dp[j] + S[j])
        for i in range(1, N+1):
            left = max(0, i - M)
            # Remove outdated elements from the front of the queues
            while que1 and que1[0] < left:
                que1.popleft()
            while que2 and que2[0] < left:
                que2.popleft()
            # Add j = i-1 to the queues if valid
            j = i - 1
            if j >= 0 and prev_dp[j] != -float('inf'):
                val1 = prev_dp[j] - S[j]
                # Maintain que1 in decreasing order of val1
                while que1 and (prev_dp[que1[-1]] - S[que1[-1]] <= val1):
                    que1.pop()
                que1.append(j)
                val2 = prev_dp[j] + S[j]
                # Maintain que2 in decreasing order of val2
                while que2 and (prev_dp[que2[-1]] + S[que2[-1]] <= val2):
                    que2.pop()
                que2.append(j)
            # Calculate the maximum possible value for current i and k
            max_val = -float('inf')
            if que1:
                current_val = (prev_dp[que1[0]] - S[que1[0]]) + S[i]
                if current_val > max_val:
                    max_val = current_val
            if que2:
                current_val = (prev_dp[que2[0]] + S[que2[0]]) - S[i]
                if current_val > max_val:
                    max_val = current_val
            if max_val != -float('inf'):
                curr_dp[i] = max_val
        prev_dp = curr_dp
    print(prev_dp[N])
if __name__ == '__main__':
    main()
            
            
            
        
            
lam6er