Overview

This problem asks us to find the sum of difficulty levels of books that Aoki can read with his viewing permits. The key insight is that when there are multiple permits, we only need to consider the one with the highest permission level.

Analysis

Key Insight

A viewing permit with permission level \(L_j\) allows viewing of “all books with difficulty at most \(L_j\)”.

For example, suppose Aoki has the following two viewing permits: - Permit A: Permission level 5 (can read books with difficulty 1, 2, 3, 4, 5) - Permit B: Permission level 10 (can read books with difficulty 1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

In this case, all books readable with Permit A are also readable with Permit B. In other words, Permit A is completely contained within Permit B.

Therefore, considering only the permit with the maximum permission level determines the set of readable books.

Problem with Naive Approach

If we enumerate readable books for each permit and take the union of sets, the time complexity becomes \(O(K \times N)\). When both \(K\) and \(N\) are \(2 \times 10^5\), this requires \(4 \times 10^{10}\) operations, which may cause TLE.

Solution

Based on the insight above, we can solve this in two steps: 1. Find the maximum permission level among the permits Aoki has 2. Sum up the difficulty levels of books whose difficulty is at most this maximum level

Algorithm

1. Read input

   • Array \(P\) of book difficulty levels
   • Array \(L\) of permit permission levels
   • Array \(T\) of permit numbers that Aoki has

2. Find maximum permission level

   • For each number \(t\) in \(T\), check \(L[t-1]\) (0-indexed)
   • Let max_level be the maximum among these values

3. Calculate the sum

   • For each book difficulty \(p\), add it to the sum if \(p \leq\) max_level

4. Output the result

Concrete Example

Input example: - Book difficulties: \(P = [3, 7, 2, 9, 5]\) - Permission levels: \(L = [4, 8, 6]\) - Permits held: \(T = [1, 3]\) (numbers 1 and 3)

Processing: - Permission level of permit 1: \(L[0] = 4\) - Permission level of permit 3: \(L[2] = 6\) - Maximum permission level: \(\max(4, 6) = 6\) - Books with difficulty 6 or less: 3, 2, 5 (books with difficulty 7, 9 cannot be read) - Sum: \(3 + 2 + 5 = 10\)

Complexity

• Time complexity: \(O(K + N)\)
  • Finding maximum permission level: \(O(K)\)
  • Summing book difficulties: \(O(N)\)
• Space complexity: \(O(N + M + K)\)
  • For storing arrays \(P\), \(L\), \(T\)

Implementation Notes

• Converting between 1-indexed and 0-indexed: Since permit numbers \(T_k\) start from 1, you need to use L[t - 1] when accessing array \(L\).
• Watch out for overflow: With maximum difficulty of \(10^9\) and up to \(2 \times 10^5\) books, the sum can reach approximately \(2 \times 10^{14}\). This is not a problem in Python since integers don’t overflow, but in other languages you need to use 64-bit integer types.

Source Code

def main():
    import sys
    input = sys.stdin.readline
    
    N, M, K = map(int, input().split())
    P = list(map(int, input().split()))
    L = list(map(int, input().split()))
    T = list(map(int, input().split()))
    
    # Find the maximum permission level among the permits Aoki has
    max_level = 0
    for t in T:
        max_level = max(max_level, L[t - 1])
    
    # Calculate the sum of difficulty levels for books with difficulty at most max_level
    total = 0
    for p in P:
        if p <= max_level:
            total += p
    
    print(total)

if __name__ == "__main__":
    main()

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