P2106 confidential espionage (computer rookie experience to be a master psychology)

Confidential Intelligence P2106

Problem Description HY is very fond of chatting with GJQ, while others are still struggling in the world of OI. To avoid disturbing other students, they exchange encrypted messages. The plaintext used for communication is a non-empty sequence made up of 0s and 1s. Additionally, it includes several password letters, each representing a different binary string. For example, the ciphertext "011a0bf00a01" represents an encrypted message where each letter corresponds to a unique binary string. The key to decrypting this cipher lies in determining what each password letter stands for. After long-term statistical analysis, we now know the fixed length of each password. Now, students who have experienced the pain of encryption have intercepted two ciphertexts, S1 and S2, and know that S1 equals S2, meaning both represent the same plaintext. Your task is to determine how many possible plaintexts exist based on these two ciphertexts. Input Format The input includes two ciphertext strings, followed by a number indicating the number of password letters, and then a list of those letters along with their corresponding lengths. Output Format Output M, which indicates the number of possible plaintexts. Sample Input 100ad1 Cc1 4 a 2 d 3 c 4 b 50 Sample Output 2 Explanation The length of the plaintext is ≤ 10000, and there is no need for high-precision calculations. This problem is actually quite straightforward. The idea is to merge positions that are the same and group them accordingly. Positions that are determined to be 0 are grouped together, as are those determined to be 1. Any positions that cannot be determined are counted, and the total number of possibilities is calculated as 2 raised to the power of the number of undetermined positions. However, if there is a conflict—such as a position being both 0 and 1—then there is no valid solution. The code uses a union-find (disjoint set) structure to manage merging of positions and ensures that no contradictions occur. If the sets containing 0 and 1 are not connected, the answer is computed as 2 raised to the number of undetermined positions. Otherwise, the answer is zero. This approach effectively handles the constraints and provides the correct number of possible plaintexts based on the given conditions.

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