Overview

This problem asks us to sequentially consider \(N\) movement plans for Takahashi on a number line, and maximize his final coordinate while avoiding collisions with barricades. Since the total sum of absolute movement distances is relatively small at \(5 \times 10^4\), the range of reachable coordinates is limited. By leveraging this observation, we can solve the problem using a fast dynamic programming (DP) approach with Bitset operations.

Analysis

1. Narrowing Down the Coordinate Range

Looking at the constraints, we have \(\sum |D_i| \leq 5 \times 10^4\). Since Takahashi starts at coordinate \(0\), no matter how he moves, the final coordinate \(s\) will always fall within the range \(-50000 \leq s \leq 50000\). Barricades \(X_j\) outside this range can be safely ignored.

2. Basic DP Formulation

Consider the following DP: - \(dp[i][s]:\) whether it is possible to be at coordinate \(s\) after considering the first \(i\) plans (boolean value)

For each plan \(D_i\), we check whether there is a barricade when moving from the current coordinate \(s\) to \(s + D_i\). If the move is possible, we set \(dp[i][s+D_i]\) to true. Additionally, when skipping a plan, \(dp[i][s]\) is always carried over.

However, the number of states is \(N \times (2 \times \sum |D_i|) \approx 5 \cdot 10^4 \times 10^5 = 5 \cdot 10^9\), which is too large for a naive DP to fit within the time limit.

3. Speedup Using Bitsets

By managing the information “whether coordinate \(s\) is reachable” as a bit sequence (a large integer in Python), we can update multiple coordinates simultaneously.

• reachable: the \(k\)-th bit is \(1\) if coordinate \(k - S\) is reachable
• can_move[d]: the corresponding bit is \(1\) if moving from coordinate \(s\) to \(s+d\) does not collide with a barricade

The transition at each step is as follows (for \(d > 0\)): reachable |= (reachable & can_move[d]) << d This allows us to compute transitions from all reachable coordinates at once.

4. Fast Barricade Collision Detection

When moving a distance \(d\) from coordinate \(s\), we need to quickly determine whether there is a barricade in the interval \([s, s+d]\) (or \([s+d, s]\)). This can be rephrased as: “Is coordinate \(s\) contained in the range from \(-d\) to \(0\) relative to some barricade coordinate \(X_j\)?”

Specifically, we prepare bad_bits where the bit corresponding to each barricade coordinate is set to \(1\), then take the OR (union) of bad_bits shifted by \(0, 1, \dots, |d|\) to efficiently compute “the set of coordinates from which moving would cause a collision.” This union can be computed in \(O(\log |d|)\) shift operations using a technique similar to doubling (Binary Lifting).

Algorithm

1. Preparation: Set the coordinate offset \(S = \sum |D_i|\), and map coordinate \(x\) to bit position \(x + S\).
2. Extract Barricades: Create a bit sequence bad_bits recording the positions of barricades within the range \([-S, S]\).
3. Precompute Movement Feasibility: For each distinct \(D_i\) appearing in the plans, compute the set of coordinates from which movement can start without collision, can_move[D_i].
   • When \(d > 0\): Take the negation of bad_bits | (bad_bits >> 1) | ... | (bad_bits >> d).
   • When \(d < 0\): Take the negation of bad_bits | (bad_bits << 1) | ... | (bad_bits << |d|).
4. Execute DP:
   • reachable = 1 << S (initial coordinate 0)
   • For each \(D_i\), update as reachable |= (reachable & can_move[D_i]) << D_i.
5. Answer: Find the highest set bit in reachable, subtract the offset \(S\), and output the result.

Complexity

• Time Complexity: \(O\left(\frac{N \cdot S}{w} + \text{distinct}(D) \cdot \frac{S}{w} \log S\right)\)
  • \(S = \sum |D_i|\), \(w\) is the word size (speedup factor from bitwise operations).
  • Bitwise operations on Python’s arbitrary-precision integers are very fast, so this complexity is sufficient to pass within the time limit.
• Space Complexity: \(O(M + S)\)
  • Depends on storing the barricades and maintaining the bitsets.

Implementation Notes

• Union of Bit Shifts: When computing B | (B >> 1) | ... | (B >> k), by doubling the shift amount as \(1, 2, 4, \dots\) while taking OR, we reduce the number of operations from \(O(k)\) to \(O(\log k)\).
• Python’s int.bit_length(): This is convenient for finding the position of the leftmost (largest coordinate) \(1\) bit in the reachable bit sequence.
• Handling Negative Coordinates: To handle the negative region of the number line, we must always add a fixed value (offset) to convert to non-negative indices.

# Core image of the bitset update process
for d in D:
    if d > 0:
        # can_move[d] is the set of coordinates s from which moving to s+d does not hit a barricade
        # Take OR of the destination from applicable coordinates (<< d) and the original coordinates (skip)
        reachable |= (reachable & can_move[d]) << d
    else:
        reachable |= (reachable & can_move[d]) >> abs(d)

Source Code

import sys

# Increase recursion depth just in case
sys.setrecursionlimit(200000)

def solve():
    # Using sys.stdin.read().split() to get all input at once
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    N = int(input_data[0])
    M = int(input_data[1])
    
    # D_i are the next N elements
    D = [int(x) for x in input_data[2:2+N]]
    
    # X_j are the next M elements
    # M can be 0, so input_data[2+N:2+N+M] will be empty
    X = [int(x) for x in input_data[2+N:2+N+M]]
    
    # Sum of absolute values of D_i is at most 50,000
    S_max = 0
    for d in D:
        if d > 0:
            S_max += d
        else:
            S_max -= d
    
    # Coordinate range of reachable s is [-S_max, S_max]
    # Offset is S_max to map to non-negative bit indices [0, 2*S_max]
    num_bits = 2 * S_max + 1
    mask = (1 << num_bits) - 1
    
    # bad_bits: bit k is 1 if coordinate k-S_max is a barricade
    bad_bits = 0
    for x in X:
        if -S_max <= x <= S_max:
            bad_bits |= (1 << (x + S_max))
    
    # Pre-calculate can_move[d] for each distinct d
    # can_move[d] bit s is 1 if the move d from coordinate s is valid (no barricades hit)
    
    def get_union_shifts_right(B, k, mask):
        # Returns B | (B >> 1) | ... | (B >> k)
        # This uses a binary exponentiation-like trick to compute the union in O(log k) shifts
        length = k + 1
        res = 0
        current_B = B
        shift_so_far = 0
        for i in range(length.bit_length()):
            if (length >> i) & 1:
                res |= (current_B >> shift_so_far)
                shift_so_far += (1 << i)
            current_B |= (current_B >> (1 << i))
            current_B &= mask
            res &= mask
        return res

    def get_union_shifts_left(B, k, mask):
        # Returns B | (B << 1) | ... | (B << k)
        length = k + 1
        res = 0
        current_B = B
        shift_so_far = 0
        for i in range(length.bit_length()):
            if (length >> i) & 1:
                res |= (current_B << shift_so_far)
                shift_so_far += (1 << i)
            current_B |= (current_B << (1 << i))
            current_B &= mask
            res &= mask
        return res

    can_move = {}
    distinct_D = set(D)
    for d in distinct_D:
        if d > 0:
            # Plan d > 0 from s is blocked if any X_j is in [s, s+d]
            # This is equivalent to s being in [X_j-d, X_j]
            # So bit s+S_max is invalid if any bit (s+S_max) is in [X_j-d+S_max, X_j+S_max]
            invalid_D = get_union_shifts_right(bad_bits, d, mask)
        else:
            # Plan d < 0 from s is blocked if any X_j is in [s+d, s]
            # This is equivalent to s being in [X_j, X_j-d] (where -d = abs(d))
            # So bit s+S_max is invalid if any bit (s+S_max) is in [X_j+S_max, X_j+abs(d)+S_max]
            invalid_D = get_union_shifts_left(bad_bits, -d, mask)
        can_move[d] = mask & ~invalid_D
    
    # DP with bitset
    # reachable is a bitset where bit k is 1 if coordinate k-S_max is reachable
    reachable = 1 << S_max
    for d in D:
        if d > 0:
            # If no collision, we can choose to execute the plan or skip it
            # Executing moves from s to s+d, skipping stays at s
            # Bit s+S_max moves to (s+d)+S_max, which is a left shift by d
            reachable |= (reachable & can_move[d]) << d
        else:
            # Executing moves from s to s+d (d is negative), which is a right shift by abs(d)
            reachable |= (reachable & can_move[d]) >> (-d)
        # We must mask reachable to stay within bounds, although it's naturally bounded
        reachable &= mask
    
    # The answer is the maximum coordinate s such that bit s+S_max is set
    # reachable.bit_length() - 1 gives the highest bit index
    print(reachable.bit_length() - 1 - S_max)

if __name__ == '__main__':
    solve()

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This editorial was generated by gemini-3-flash-thinking.