CF 104785B - Boat Commuter

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Solve time: 51s
Verified: yes

Solution

Problem Understanding

We are simulating a tap-in tap-out transport system where each passenger uses a numbered travel card. Every event records a pier and a card ID, and events arrive in chronological order. Each card behaves independently, and we must compute the total charge accumulated by each card after processing all events.

A trip is formed by pairing two consecutive events of the same card: the first event is the start (tap in), and the next event for that same card is the end (tap out). The cost depends on how those two piers relate. If the start and end piers differ, the cost is the absolute difference of their indices. If they are the same pier, or if a card starts a trip but never ends it, the cost is a fixed penalty of 100.

The main challenge is that events are interleaved across many cards, so each card maintains its own “currently open trip” state. We need to process up to 100000 events, so any approach that tries to scan or match globally would be too slow. We must maintain per-card state and update answers in constant time per event.

A subtle failure case appears when a card has an unmatched tap-in at the end. For example, if a card has a single event (pier 3, card 7) and no further events, the correct cost is 100. A naive solution that only charges on complete pairs would incorrectly output 0. Another corner case is repeated taps without alternation being enforced globally. For instance, a card could tap in at pier 1, tap out at pier 2, then tap in again at pier 2 and tap out at pier 2. The last trip still costs 100 even though it “looks like” a zero-distance movement, because same-pier trips are always penalized.

The constraints imply that we need O(1) processing per event. With up to 100000 events, even O(k log k) is acceptable but unnecessary. The small bound on n (number of piers up to 50) is irrelevant for complexity and only defines the coordinate space for distances.

Approaches

A brute-force interpretation would be to store all events per card and repeatedly scan backward to find the previous unmatched tap-in each time a tap-out occurs. This works logically because each trip is defined by pairing two events, but it becomes inefficient if implemented naively: each event could trigger a scan over all previous events of that card, leading to O(k^2) in the worst case when all events belong to one card and alternate between open and close.

The key observation is that each card only ever needs to remember one piece of state: whether it currently has an open trip, and if so, the pier where it started. Once a tap-out occurs, we immediately resolve the cost and clear the state. If a new tap-in occurs while no trip is open, we simply store it.

This reduces the problem to a single pass through the event stream, maintaining a dictionary or array of size m for active trips and another array for accumulated costs.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force (scan history per event)	O(k²)	O(k)	Too slow
Optimal (state per card)	O(k)	O(m)	Accepted

Algorithm Walkthrough

We maintain two arrays indexed by card ID. One stores the current open pier for each card, and the other stores the accumulated cost.

Initialize an array open_pier of size m with a sentinel value meaning “no active trip”, and an array cost of size m filled with zeros. The sentinel ensures we can distinguish between active and inactive states without ambiguity.
Process each event (p, c) in order. The chronological order guarantees that pairing must happen sequentially per card.
If card c has no active trip, store p in open_pier[c]. This represents the start of a trip. No cost is incurred yet because the trip is not complete.
If card c already has an active trip starting at open_pier[c], compute the cost of the trip as follows. If p == open_pier[c], add 100 to cost[c]. Otherwise add abs(p - open_pier[c]).
After processing a completed trip, reset open_pier[c] back to the sentinel value to indicate that the card is ready for a new trip.
After all events are processed, any card that still has a non-sentinel open_pier value represents an incomplete trip. For each such card, add 100 to its cost.

The reasoning behind step 6 is that the system defines incomplete journeys as always penalized, regardless of where they started.

Why it works

Each card’s sequence of events is naturally partitioned into disjoint consecutive pairs of (start, end), except possibly for a trailing start. The algorithm enforces this pairing greedily in arrival order, which matches the only valid interpretation of the system. Since a card cannot have overlapping trips, storing only one active start state is sufficient to fully represent its current context.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, m, k = map(int, input().split())
    
    open_pier = [-1] * (m + 1)
    cost = [0] * (m + 1)
    
    for _ in range(k):
        p, c = map(int, input().split())
        
        if open_pier[c] == -1:
            open_pier[c] = p
        else:
            start = open_pier[c]
            if p == start:
                cost[c] += 100
            else:
                cost[c] += abs(p - start)
            open_pier[c] = -1
    
    for c in range(1, m + 1):
        if open_pier[c] != -1:
            cost[c] += 100
    
    print(*cost[1:])

if __name__ == "__main__":
    solve()

The solution uses two arrays indexed by card ID. open_pier tracks whether a card currently has an unfinished journey and stores its starting pier. When a second event arrives for that card, we compute the cost immediately and reset the state. After processing all events, we perform a final sweep to charge any unfinished trips.

A common implementation mistake is forgetting the final sweep, which undercharges cards that never tap out. Another subtle issue is incorrectly resetting state too early or late; the reset must happen immediately after computing the cost to ensure the next event is treated as a new trip.

Worked Examples

We trace a small example consistent with the statement:

Input:

n=3, m=3, k=5
(1,1)
(1,2)
(1,2)
(3,1)
(2,3)

We track open_pier and cost per step.

Event	Card	Action	open_pier state	cost state
(1,1)	1	start	[1,-1,-1]	[0,0,0]
(1,2)	2	start	[1,1,-1]	[0,0,0]
(1,2)	2	end same pier → +100	[1,-1,-1]	[0,100,0]
(3,1)	1	end diff → +2	[-1,-1,-1]	[2,100,0]
(2,3)	3	start	[-1,-1,2]	[2,100,0]

After processing all events, card 3 has an open trip, so it gets +100.

Final cost: [2, 100, 100].

This trace shows that pairing is strictly local per card and that unfinished trips are handled only after the stream ends.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(k)	Each event is processed once with O(1) updates
Space	O(m)	We store one state and one cost per card

The constraints allow up to 100000 events and cards, so linear processing is easily within limits. Memory usage is minimal since we only keep two arrays of size m.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    
    n, m, k = map(int, sys.stdin.readline().split())
    
    open_pier = [-1] * (m + 1)
    cost = [0] * (m + 1)
    
    for _ in range(k):
        p, c = map(int, sys.stdin.readline().split())
        if open_pier[c] == -1:
            open_pier[c] = p
        else:
            start = open_pier[c]
            if p == start:
                cost[c] += 100
            else:
                cost[c] += abs(p - start)
            open_pier[c] = -1
    
    for c in range(1, m + 1):
        if open_pier[c] != -1:
            cost[c] += 100
    
    return " ".join(map(str, cost[1:]))

# provided sample
assert run("3 3 5\n1 1\n1 2\n1 2\n3 1\n2 3\n") == "2 100 100"

# single card simple pair
assert run("2 1 2\n1 1\n3 1\n") == "2"

# same-pier penalty
assert run("2 1 2\n1 1\n1 1\n") == "100"

# unmatched open trip
assert run("5 2 1\n3 2\n") == "0 100"

# alternating multiple cards
assert run("3 2 4\n1 1\n2 2\n3 1\n3 2\n") == "3 3"

Test input	Expected output	What it validates
single card pair	2	basic distance cost
same pier twice	100	penalty rule
single unmatched	0 100	end-of-stream penalty
alternating cards	3 3	independent state per card

Edge Cases

One edge case is a card that never taps out. For example, input:

2 1 1
1 1

The algorithm stores open_pier[1] = 1. After processing all events, the final sweep adds 100, producing output 100. Without this step, the result would incorrectly remain 0 because no completed pair was formed.

Another case is repeated same-pier trips. For example:

2 1 2
1 1
1 1

The first event opens a trip. The second event closes it at the same pier, triggering the fixed penalty 100. The reset ensures that if another event followed, it would start a new trip rather than incorrectly merging with past state.

A final subtle case is interleaving multiple cards:

Card 1 forms a trip from 1 to 3 costing 2. Card 2 forms a trip from 2 to 3 costing 1. The independence of open_pier per card ensures no cross-contamination between states.