Overview

Given \(N\) types of strings (each with a limited stock), select some strings so that the total length is exactly \(W\), while minimizing the number of strings used. This is a variant of the “Bounded Knapsack Problem.”

Analysis

Essence of the Problem

Each type of string \(i\) has length \(L_i\) and can be used up to \(C_i\) times. We want to make the total length exactly \(W\) while minimizing the number of strings used. This is a classic Bounded Knapsack Problem.

Naive Approaches and Their Issues

Approach 1: Treat each string individually We could treat each type \(i\) as \(C_i\) individual items and solve a 0-1 knapsack. However, since \(C_i\) can be up to \(10000\) and \(N\) up to \(100\), the total number of items can reach \(100 \times 10000 = 10^6\). Since each item requires \(O(W)\) updates, the total complexity becomes \(O(10^6 \times 50000) = O(5 \times 10^{10})\), which is far too slow.

Approach 2: Unbounded knapsack style If we ignore the stock limits and allow unlimited use, it becomes an unbounded knapsack problem. However, this does not correctly solve our problem since stock limits exist.

Solution: Binary Splitting

We use a technique that splits the quantity limit \(C_i\) into groups corresponding to each digit in binary representation.

For example, when \(C_i = 13\), we split as \(13 = 1 + 2 + 4 + 6\) (\(1, 2, 4\) are powers of \(2\), and \(6\) is the remainder). By combining these groups, we can represent any number of strings from \(0\) to \(13\).

This compresses \(C_i\) items into \(O(\log C_i)\) groups, and we only need to perform 0-1 knapsack updates for each group.

Algorithm

1. Initialize the DP array: Let \(dp[w]\) be “the minimum number of strings such that the total length is exactly \(w\).” Set \(dp[0] = 0\) and all others to \(\infty\).

2. Binary split each type of string: Split the stock \(C_i\) of type \(i\) into groups of \(1, 2, 4, 8, \ldots\). For example, if \(C_i = 13\), split into \(\{1, 2, 4, 6\}\).

3. 0-1 knapsack update for each group: If the group contains \(c\) strings, treat it as a single item with “weight \(L_i \times c\) and cost \(c\).” Loop \(w\) from \(W\) down to \(L_i \times c\) and update: $\(dp[w] = \min(dp[w],\ dp[w - L_i \times c] + c)\)$

4. Output the result: If \(dp[W]\) is \(\infty\), output \(-1\); otherwise, output \(dp[W]\).

Complexity

• Time complexity: \(O(N \cdot W \cdot \log C_{\max})\)
  • Each type has \(O(\log C_i)\) groups, and each group update takes \(O(W)\)
  • With \(N = 100\), \(W = 50000\), \(\log_2(10000) \approx 14\), this is approximately \(7 \times 10^7\)
• Space complexity: \(O(W)\) (size of the DP array)

Implementation Notes

• Implementing binary splitting: Start with \(k = 1, 2, 4, \ldots\), doubling each time, and take min(k, remaining) from the remaining count remaining to form each group. The key point is that the last group may be a non-power-of-\(2\) remainder.

• Reverse loop for 0-1 knapsack: Since each group is treated as “one item,” we update \(w\) from large to small to prevent reusing the same group multiple times.

• Skip cases where weight = L * count exceeds \(W\) to avoid unnecessary computation.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    W = int(input_data[idx]); idx += 1
    
    items = []
    for i in range(N):
        L = int(input_data[idx]); idx += 1
        C = int(input_data[idx]); idx += 1
        items.append((L, C))
    
    INF = float('inf')
    # dp[w] = minimum number of strings to make exactly w cm
    dp = [INF] * (W + 1)
    dp[0] = 0
    
    # Bounded knapsack with binary splitting
    for L, C in items:
        # Split C into powers of 2
        groups = []
        remaining = C
        k = 1
        while remaining > 0:
            take = min(k, remaining)
            groups.append(take)
            remaining -= take
            k *= 2
        
        for count in groups:
            # Each group: using 'count' strings of length L
            # This is like a 0-1 knapsack item with weight = L*count, value = count
            weight = L * count
            if weight > W:
                continue
            for w in range(W, weight - 1, -1):
                if dp[w - weight] + count < dp[w]:
                    dp[w] = dp[w - weight] + count
    
    if dp[W] == INF:
        print(-1)
    else:
        print(dp[W])

solve()

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