Overview

This problem asks us to select from \(N\) given requests (items) such that “the total weight is at most \(S\)” and “the total net profit is at least \(T\)”, and find the minimum number of selected items under these conditions.

As an application of the classic knapsack problem, this can be solved using dynamic programming (DP).

Analysis

First, we compute the net profit of each request as \(V_i = P_i - C_i\). What we want to achieve is the following two conditions: 1. \(\sum W_i \leq S\) (weight constraint) 2. \(\sum V_i \geq T\) (profit target)

Among all selections satisfying these conditions, we want to minimize the number of selected items \(k\).

Why isn’t a simple knapsack sufficient?

The standard knapsack problem is about “maximizing value within a weight limit,” but here we need to “minimize the number of items while also meeting a value (profit) target.” Therefore, we need to incorporate the number of selected items into the DP “state.”

Efficient State Design

Consider the following DP table: dp[k][w]: The maximum net profit when exactly \(k\) requests are selected and the total weight is exactly \(w\).

With this definition, after processing all items, we can find the answer by searching for the smallest \(k\) such that dp[k][w] >= T for some valid \(w\).

Additionally, requests with net profit \(V_i \leq 0\) do not need to be considered, since selecting them only increases the count \(k\) without increasing (or even decreasing) the profit, which is unhelpful for minimizing the number of items.

Algorithm

We solve this using dynamic programming (DP).

1. Preprocessing:
   • Compute the net profit \(V_i = P_i - C_i\) for each request, and extract only those with \(V_i > 0\).
2. DP Table Definition:
   • dp[k][w]: The maximum net profit when \(k\) items are selected and the total weight is \(w\).
   • Initialize all values to -1 (unreachable), with dp[0][0] = 0.
3. Transitions: For each request \((V_i, W_i)\), update the current DP table.
   • If we can already reach weight \(w\) with \(k\) items (dp[k][w] != -1), then by adding this request, we can create a state with \(k+1\) items and weight \(w + W_i\).
   • dp[k+1][w + W_i] = max(current value, dp[k][w] + V_i)
4. Finding the Answer:
   • Check in order \(k = 1, 2, \dots, N\).
   • If for any weight \(w \in [0, S]\), dp[k][w] >= T, then that \(k\) is the minimum number of items.
   • If no valid \(k\) is found, output -1.

Complexity

• Time Complexity: \(O(N^2 S)\)
  • A triple loop (or equivalent processing) occurs over the number of requests \(N\), the maximum number of selected items \(N\), and the capacity \(S\).
  • Given the constraints \(N=50, S=10^4\), this results in approximately \(50 \times 50 \times 10^4 = 2.5 \times 10^7\) operations, which comfortably fits within the time limit.
• Space Complexity: \(O(N S)\)
  • This is for maintaining the dp[N+1][S+1] table.

Implementation Notes

• Order of count updates: To avoid selecting the same request more than once, the loop over the count \(k\) must iterate from larger values to smaller, or a separate table for updates must be prepared.

• Optimization (reachable weight list): In languages like Python, scanning all values of \(w\) can be slow. In the code, a reachable_weights list is used to only update weights that are actually reachable, thereby improving performance.

• Only consider requests with positive net profit: Requests with \(P_i - C_i \leq 0\) can be excluded, since there is no benefit to increasing the count for achieving the profit target \(T \geq 1\).

  Source Code

import sys

def solve():
    # Read all input data at once for speed
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    # Parse N, S, T from the first three elements
    N = int(input_data[0])
    S = int(input_data[1])
    T = int(input_data[2])
    
    # Parse each request's reward, cost, and weight
    items = []
    for i in range(3, 3 + 3 * N, 3):
        p = int(input_data[i])
        c = int(input_data[i+1])
        w = int(input_data[i+2])
        # Only consider requests with positive net profit
        # Requests with net profit <= 0 will never decrease the minimum count of items
        if p - c > 0:
            items.append((p - c, w))
    
    # If no items have positive profit, and T >= 1, no solution exists
    if not items:
        print("-1")
        return

    # dp[k][w] stores the maximum profit using exactly k items and exactly weight w
    # Initialize with -1 to indicate that the weight is not yet reachable
    dp = [[-1] * (S + 1) for _ in range(N + 1)]
    dp[0][0] = 0
    
    # reachable_weights[k] stores a list of unique weights reachable with exactly k items
    reachable_weights = [[] for _ in range(N + 1)]
    reachable_weights[0].append(0)
    
    # num_items_processed tracks how many requests we've considered so far
    num_items_processed = 0
    for v, w_i in items:
        # Iterate backwards through the number of items to ensure each request is used at most once
        for k in range(num_items_processed, -1, -1):
            dk = dp[k]
            dk1 = dp[k+1]
            rw = reachable_weights[k]
            rw1 = reachable_weights[k+1]
            for w in rw:
                nw = w + w_i
                # Check if the total weight is within the capacity S
                if nw <= S:
                    nv = dk[w] + v
                    # If this weight hasn't been reached with k+1 items yet
                    if dk1[nw] == -1:
                        dk1[nw] = nv
                        rw1.append(nw)
                    # If it has been reached, update it if the new profit is higher
                    elif nv > dk1[nw]:
                        dk1[nw] = nv
        
        # Increment the count of processed items, capped at N
        if num_items_processed < N:
            num_items_processed += 1

    # Find the minimum k such that there exists a weight w with profit at least T
    for k in range(1, N + 1):
        dk = dp[k]
        for w in reachable_weights[k]:
            if dk[w] >= T:
                print(k)
                return
    
    # If no such k is found, print -1
    print("-1")

if __name__ == "__main__":
    solve()

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