Overview

This problem asks for the minimum number of round trips needed to transport \(N\) items using an elevator with a weight capacity of \(C\). This is a type of “bin packing problem” and can be solved by combining bit exhaustive search with DP.

Analysis

Essence of the Problem

• We want to divide items into groups where the total weight is at most \(C\)
• We want to minimize the number of groups (= number of round trips)

Naive Approach

If we try all possible divisions, the number of ways to partition \(N\) items is called the “Bell number,” which for \(N = 15\) is approximately \(10^9\) ways, resulting in TLE.

Key Insight

Notice the constraint \(N \leq 15\). Since \(2^{15} = 32768\), we can represent item selections (subsets) using bitmasks.

For example, when \(N = 3\): - 101 (binary) = state where items 1 and 3 are selected - 111 = state where all items are selected

Solution Strategy

We use DP with “which items have been transported” as the state.

• dp[mask] = minimum number of round trips needed to transport all items represented by mask

The transition is “select a combination of items with total weight at most \(C\) from the remaining items and transport them in one trip.”

Algorithm

1. Preprocessing: Calculate the total weight for all \(2^N\) subsets

2. DP Initialization: dp[0] = 0 (0 trips needed when nothing has been transported), others are \(\infty\)

3. DP Transition: For each state mask

   • Calculate remaining items remaining = full_mask XOR mask
   • Enumerate subsets sub of remaining
   • If the weight of sub is at most \(C\), then dp[mask | sub] = min(dp[mask | sub], dp[mask] + 1)

4. Answer: dp[(1 << N) - 1] (state where all items have been transported)

Subset Enumeration Technique

All non-empty subsets of a set remaining can be enumerated in \(O(2^{|remaining|})\) with the following loop:

sub = remaining
while sub > 0:
    # process sub
    sub = (sub - 1) & remaining

Complexity

• Time Complexity: \(O(3^N)\)

  • The number of states is \(2^N\), and the total number of subsets transitioned from each state is \(\sum_{k=0}^{N} \binom{N}{k} 2^{N-k} = 3^N\)

• Space Complexity: \(O(2^N)\)

  • \(2^N\) elements each for the dp array and subset_weight array

When \(N = 15\), \(3^{15} \approx 1.4 \times 10^7\), which is fast enough.

Implementation Notes

1. Precompute weights: Computing subset weights each time is slow, so precompute subset_weight[mask]

2. Subset enumeration formula: sub = (sub - 1) & remaining is a common competitive programming technique. It efficiently enumerates only subsets of positions where bits are set in remaining

3. Handling INF: Skip unreachable states with if dp[mask] == INF: continue to avoid transitioning from them

Source Code

def solve():
    N, C = map(int, input().split())
    W = list(map(int, input().split()))
    
    # Calculate the total weight of items for each subset
    subset_weight = [0] * (1 << N)
    for mask in range(1 << N):
        total = 0
        for i in range(N):
            if mask & (1 << i):
                total += W[i]
        subset_weight[mask] = total
    
    # Enumerate valid subsets (weight at most C)
    valid_subsets = []
    for mask in range(1 << N):
        if subset_weight[mask] <= C:
            valid_subsets.append(mask)
    
    # dp[mask] = minimum number of round trips needed to transport all items represented by mask
    INF = float('inf')
    dp = [INF] * (1 << N)
    dp[0] = 0
    
    full_mask = (1 << N) - 1
    
    for mask in range(1 << N):
        if dp[mask] == INF:
            continue
        
        # Remaining items
        remaining = full_mask ^ mask
        
        # Enumerate subsets of remaining items
        sub = remaining
        while sub > 0:
            if subset_weight[sub] <= C:
                new_mask = mask | sub
                if dp[new_mask] > dp[mask] + 1:
                    dp[new_mask] = dp[mask] + 1
            # Next subset
            sub = (sub - 1) & remaining
    
    print(dp[full_mask])

solve()

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