Overview

This problem asks us to select a set of mutually non-overlapping events and maximize the profit, which is the revenue from accepting events minus the cancellation compensation costs. Through algebraic transformation, this can be reduced to a Weighted Interval Scheduling problem.

Analysis

Transforming the Profit Formula

Let \(S\) be the set of accepted events. The profit can be expressed as:

\[|S| \times B - \sum_{i \notin S} C_i\]

Here, we transform the total compensation cost for rejected events:

\[\sum_{i \notin S} C_i = \sum_{i=1}^{N} C_i - \sum_{i \in S} C_i\]

Substituting this into the profit formula:

\[|S| \times B - \left(\sum_{i=1}^{N} C_i - \sum_{i \in S} C_i\right) = \sum_{i \in S}(B + C_i) - \sum_{i=1}^{N} C_i\]

Since \(\sum_{i=1}^{N} C_i\) is a constant, maximizing the profit is equivalent to maximizing \(\sum_{i \in S}(B + C_i)\).

Reducing the Problem

Assign each event \(i\) a weight \(w_i = B + C_i\), and the problem becomes maximizing the total weight of a set of mutually non-overlapping intervals. This is the classic Weighted Interval Scheduling problem.

Issues with the Naive Approach

Enumerating all subsets of events takes \(O(2^N)\), which is far too slow for \(N \leq 2 \times 10^5\). This can be solved efficiently by combining dynamic programming with binary search.

Algorithm

1. Preprocessing: Sort all events in ascending order of end time \(R_i\).
2. DP Definition: Let \(dp[i]\) be “the maximum total weight achievable when only considering the first \(i\) events after sorting.” \(dp[0] = 0\).
3. Transition: For the \(i\)-th event (0-indexed):
   • Don’t select it: \(dp[i+1] = dp[i]\)
   • Select it: Find the latest event that doesn’t overlap with this event. Specifically, use binary search to find the largest \(j\) satisfying \(R_j \leq L_i\), and set \(dp[i+1] = w_i + dp[\text{number of such events}]\).
   • Take the \(\max\) of both cases.
4. Computing the Answer: Output \(dp[N] - \sum_{i=1}^{N} C_i\).

Concrete Example

For example, if the events are \([1, 3), [2, 5), [4, 6)\) with \(B = 10\), \(C = [3, 2, 5]\): - Weights: \(w = [13, 12, 15]\) - Already sorted by end time (in this order) - When selecting event \([4, 6)\), the event satisfying \(R_j \leq 4\) is \([1, 3)\), so we can select both \([1,3)\) and \([4,6)\) simultaneously, with total weight \(13 + 15 = 28\) - Profit: \(28 - (3+2+5) = 18\)

Complexity

• Time complexity: \(O(N \log N)\) (\(O(N \log N)\) for sorting, \(O(\log N)\) for binary search per event)
• Space complexity: \(O(N)\) (for the DP array and event information storage)

Implementation Notes

• Handling half-open intervals: The condition for intervals \([L_i, R_i)\) to not overlap is \(R_j \leq L_i\). Using bisect_right(ends, L_i, 0, i) correctly computes the number of events whose end time is at most \(L_i\). When endpoints coincide (\(R_j = L_i\)), the intervals do not overlap, so bisect_right is the appropriate choice.

• When the answer is negative: Since compensation costs are incurred even if all events are rejected, the answer can be negative. No special case handling is needed — the formula \(dp[N] - \text{total\_C}\) directly returns the correct answer.

• Overflow: In Python, this is not a concern due to arbitrary-precision integers, but when implementing in C++ or similar languages, long long is required (\(B + C_i\) can be up to \(2 \times 10^9\), and with \(N\) up to \(2 \times 10^5\), the total can reach approximately \(4 \times 10^{14}\)).

  Source Code

import sys
from bisect import bisect_right

def main():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    B = int(input_data[idx]); idx += 1
    
    events = []
    total_C = 0
    for i in range(N):
        L = int(input_data[idx]); idx += 1
        R = int(input_data[idx]); idx += 1
        C = int(input_data[idx]); idx += 1
        events.append((L, R, C))
        total_C += C
    
    # Profit = |S| * B - sum_{i not in S} C_i
    #        = |S| * B - (total_C - sum_{i in S} C_i)
    #        = sum_{i in S} (B + C_i) - total_C
    #
    # So we want to maximize sum_{i in S} (B + C_i) where S is a set of 
    # non-overlapping intervals (half-open [L_i, R_i)).
    # This is a weighted interval scheduling problem.
    
    # Weight of each event
    weights = [(events[i][0], events[i][1], B + events[i][2]) for i in range(N)]
    
    # Sort by end time
    weights.sort(key=lambda x: x[1])
    
    ends = [w[1] for w in weights]
    
    # dp[i] = max weight using events from the first i events (0-indexed: events 0..i-1)
    # dp[0] = 0
    # dp[i] = max(dp[i-1], w_i + dp[j]) where j is the largest index such that ends[j] <= L_i
    
    dp = [0] * (N + 1)
    
    for i in range(N):
        L_i, R_i, w_i = weights[i]
        # Find largest j such that ends[j] <= L_i, i.e., j+1 events considered
        # ends is sorted. We want rightmost index where ends[idx] <= L_i
        # bisect_right(ends, L_i) gives the number of elements <= L_i... no.
        # bisect_right(ends, L_i) gives first index where ends[index] > L_i
        # So the number of events (among first i events, indices 0..i-1) with end <= L_i
        # But we only look at ends[0..i-1] since we're considering event i (0-indexed)
        # Actually, ends has all N elements. We want among indices 0..i-1.
        
        j = bisect_right(ends, L_i, 0, i)  # number of events among 0..i-1 with end <= L_i
        
        dp[i + 1] = max(dp[i], w_i + dp[j])
    
    max_weight_sum = dp[N]
    answer = max_weight_sum - total_C
    
    print(answer)

main()

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