Overview

This is a problem where we need to determine whether there exists a way to select requests such that the total weight is at most \(S\) and the total net profit is at least \(T\), and if so, find the minimum number of requests to select. We solve it using dynamic programming (DP) as a variant of the 0/1 knapsack problem.

Analysis

Problem Formulation

Each request \(i\) has the following three values: - Net profit: \(P_i - C_i\) (can be negative) - Weight: \(W_i\)

From this, we want to find the subset with the minimum number of elements such that the total weight is at most \(S\) and the total net profit is at least \(T\).

Naive Approach

Enumerating all subsets gives \(2^N\) possibilities, and when \(N = 50\), \(2^{50} \approx 10^{15}\), which is far too slow.

Key Insight

If we can answer “when fixing the number of selected items to \(k\), what is the maximum profit achievable by selecting exactly \(k\) items with total weight at most \(S\)?”, then we can check \(k = 0, 1, 2, \ldots\) in increasing order, and the first \(k\) where the maximum profit is at least \(T\) is the answer.

This is an extension of the 0/1 knapsack problem, solvable by adding “number of selected items” to the state.

Algorithm

DP Definition

\(dp[k][w]\) = the maximum value of the total net profit when exactly \(k\) requests are selected and the total weight of the items is exactly \(w\)

• Initial state: \(dp[0][0] = 0\), all others are \(-\infty\)
• Transition: For each request \((profit_i, w_i)\), loop in reverse order just like in the 0/1 knapsack

\[dp[k+1][w + w_i] = \max(dp[k+1][w + w_i],\ dp[k][w] + profit_i)\]

The transition is only applied when \(w + w_i \leq S\). We iterate in reverse order to prevent using the same item more than once.

Finding the Answer

After processing all items, check the following for \(k = 0, 1, 2, \ldots, N\) in order:

\[\max_{0 \leq w \leq S} dp[k][w] \geq T\]

The first \(k\) that satisfies this is the answer. If no \(k\) satisfies it, output -1.

Concrete Example

For example, with \(N=3, S=10, T=5\) and requests \((profit=3, w=4), (profit=4, w=5), (profit=2, w=3)\): - \(k=1\): Maximum profit is \(4\) (only the 2nd request) → does not reach \(T=5\) - \(k=2\): \((profit=3+4=7, w=4+5=9 \leq 10)\) → \(7 \geq 5\), so the answer is \(2\)

Complexity

• Time complexity: \(O(N^2 \times S)\)
  • For each item (\(N\) items), we run a double loop over \(k\) (up to \(N\)) and \(w\) (up to \(S\))
  • When \(N = 50, S = 10000\), this is approximately \(2.5 \times 10^7\), which is sufficiently fast
• Space complexity: \(O(N \times S)\)
  • Size of the DP table

Implementation Notes

• Reverse order loop: In the 0/1 knapsack, to prevent using the same item multiple times, we iterate \(k\) and \(w\) from large to small. Forgetting this will cause the same request to be selected multiple times.

• Requests with negative net profit: Requests where \(P_i - C_i < 0\) can exist, but they are naturally handled by the DP transitions (selecting them decreases profit, so they are typically not chosen in the optimal solution).

• Managing unreachable states: By checking not to transition from states where \(dp[k][w] = -\infty\), we correctly exclude states that are actually unreachable.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    S = int(input_data[idx]); idx += 1
    T = int(input_data[idx]); idx += 1
    
    items = []
    for i in range(N):
        P = int(input_data[idx]); idx += 1
        C = int(input_data[idx]); idx += 1
        W = int(input_data[idx]); idx += 1
        profit = P - C
        items.append((profit, W))
    
    # DP: dp[w] = (max_profit, min_count_for_that_max_profit)
    # We want to find minimum count such that weight <= S and profit >= T
    # 
    # Better approach: dp[w] = dict or we do 2D DP
    # Since we want to minimize count, let's do:
    # For each number of items k (from 0 to N), check if it's achievable.
    # But that's complex too.
    #
    # Alternative: dp[w] = maximum profit achievable with weight exactly w, and for each state also track minimum items.
    # But we need to be more careful: among all subsets with weight <= S and profit >= T, minimize count.
    #
    # Let's do: dp[w][count] = max profit with total weight w using exactly count items
    # That's O(N * S * N) which is O(50 * 10000 * 50) = 25,000,000 - feasible.
    #
    # Or: for each count k from 0..N, dp_k[w] = max profit using exactly k items with weight w
    # Return first k where max over w<=S of dp_k[w] >= T
    
    # dp[w] = max profit achievable with weight exactly w, using minimum items
    # Actually let's track: dp[w] = list where dp[w] gives max profit for weight w
    # and we want min count. 
    #
    # Cleaner: dp[w] = (max_profit, min_count) isn't right because max profit with fewer items 
    # might be less than max profit with more items.
    #
    # Best approach: dp[w] = {count: max_profit} but that's expensive.
    #
    # Let me think again. We want minimum count. So iterate k from 0 to N:
    # For each k, check if there's a subset of exactly k items with weight <= S and profit >= T.
    # Use DP: after considering all items, dp[k][w] = max profit using exactly k items with weight exactly w.
    # Answer = smallest k such that max(dp[k][w] for w in 0..S) >= T.
    # Space: O(N * S) = O(50 * 10000) = 500,000. Time: O(N^2 * S) = O(50 * 50 * 10000) = 25M. Fine.
    
    NEG_INF = float('-inf')
    
    # dp[k][w] = max profit using exactly k items with total weight exactly w
    # k ranges 0..N, w ranges 0..S
    # We only care about w <= S
    
    # Initialize: dp[0][0] = 0, everything else = -inf
    # For memory efficiency, use 2D array dp[N+1][S+1]
    
    dp = [[NEG_INF] * (S + 1) for _ in range(N + 1)]
    dp[0][0] = 0
    
    for profit, w in items:
        # Traverse in reverse to avoid using same item twice (0/1 knapsack)
        for k in range(min(N, N) - 1, -1, -1):  # from N-1 down to 0
            for ww in range(S + 1):
                if dp[k][ww] == NEG_INF:
                    continue
                new_w = ww + w
                if new_w <= S:
                    new_profit = dp[k][ww] + profit
                    if new_profit > dp[k + 1][new_w]:
                        dp[k + 1][new_w] = new_profit
    
    for k in range(N + 1):
        max_profit = NEG_INF
        for ww in range(S + 1):
            if dp[k][ww] > max_profit:
                max_profit = dp[k][ww]
        if max_profit >= T:
            print(k)
            return
    
    print(-1)

solve()

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