結果
問題 | No.765 ukuku 2 |
ユーザー |
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提出日時 | 2025-06-12 17:08:14 |
言語 | PyPy3 (7.3.15) |
結果 |
TLE
|
実行時間 | - |
コード長 | 4,580 bytes |
コンパイル時間 | 391 ms |
コンパイル使用メモリ | 82,504 KB |
実行使用メモリ | 89,960 KB |
最終ジャッジ日時 | 2025-06-12 17:08:22 |
合計ジャッジ時間 | 7,082 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 28 TLE * 1 -- * 19 |
ソースコード
def longest_palindrome_after_deletion(s): n = len(s) if n == 1: return 0 # Manacher's algorithm to find all possible palindromes # Preprocess the string t = '#' + '#'.join(s) + '#' n_t = len(t) P = [0] * n_t C, R = 0, 0 max_len = 0 max_indices = [] for i in range(n_t): mirror = 2 * C - i if i < R: P[i] = min(R - i, P[mirror]) # Attempt to expand palindrome centered at i a, b = i + (1 + P[i]), i - (1 + P[i]) while a < n_t and b >= 0 and t[a] == t[b]: P[i] += 1 a += 1 b -= 1 # Update center and right boundary if palindrome centered at i expands past R if i + P[i] > R: C = i R = i + P[i] # Update max_len and record the centers that achieve this max_len if P[i] > max_len: max_len = P[i] max_indices = [i] elif P[i] == max_len: max_indices.append(i) # Convert back to original string's indices max_len = (max_len - 1) // 2 # since in t, each character is doubled # Now, for each i, compute the max palindrome length after deleting s[i] max_after_deletion = 0 # Precompute for each position the maximum palindrome length in the prefix and suffix left_max = [0] * n right_max = [0] * n # Compute left_max for i in range(n): if i == 0: left_max[i] = 1 continue current_max = left_max[i-1] for j in range(0, i+1): if s[j] == s[i] and (i - j + 1) > current_max: is_pal = True for k in range(j+1, (j+i)//2 + 1): if s[k] != s[i - (k - j)]: is_pal = False break if is_pal: current_max = i - j + 1 break left_max[i] = current_max # Compute right_max for i in range(n-1, -1, -1): if i == n-1: right_max[i] = 1 continue current_max = right_max[i+1] for j in range(n-1, i-1, -1): if s[j] == s[i] and (j - i + 1) > current_max: is_pal = True for k in range(i+1, (i+j)//2 +1): if s[k] != s[j - (k - i)]: is_pal = False break if is_pal: current_max = j - i + 1 break right_max[i] = current_max # Precompute the maximum palindrome lengths using Manacher's data # Now, for each i, compute the maximum palindrome length after deletion for i in range(n): left_part = s[:i] right_part = s[i+1:] combined = left_part + right_part current_max = 0 # Check for the maximum palindrome in left_part, right_part, and combined # Precompute the maximum of left and right max_left = left_max[i-1] if i > 0 else 0 max_right = right_max[i+1] if i < n-1 else 0 current_max = max(max_left, max_right) # Check combined # Use Manacher's algorithm to find the maximum palindrome in combined # This is a fallback method and may be too slow for large n, but for the sake of example # We'll proceed with it if len(combined) > current_max: # Compute the maximum palindrome in combined using a simple method def max_pal(s): n = len(s) max_len = 1 for i in range(n): # Odd length l, r = i, i while l >=0 and r < n and s[l] == s[r]: if r - l +1 > max_len: max_len = r - l +1 l -=1 r +=1 # Even length l, r = i, i+1 while l >=0 and r < n and s[l] == s[r]: if r - l +1 > max_len: max_len = r - l +1 l -=1 r +=1 return max_len combined_max = max_pal(combined) current_max = max(current_max, combined_max) if current_max > max_after_deletion: max_after_deletion = current_max return max_after_deletion # Read input s = input().strip() # Compute the result result = longest_palindrome_after_deletion(s) # Output the result print(result)