CF 104777M - Treasure Chest

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Solve time: 57s
Verified: yes

Solution

Problem Understanding

We are working on a one-dimensional line of integer coordinates. Monocarp starts at position 0. There is a key at position y and a treasure chest at position x. To finish the task, Monocarp must eventually be at the chest’s location and possess the key at the same time so that he can open the chest.

He can walk left or right one unit per second. Picking up the key or chest takes no time, and opening the chest also takes no time once both conditions are satisfied.

The only extra complication is that the chest is heavy. If Monocarp carries it while moving, each second spent moving with the chest in hand consumes stamina. Across the entire journey, the total time spent carrying the chest cannot exceed k. Dropping and picking up the chest does not reset this counter, so all carrying intervals accumulate.

The goal is to compute the minimum total walking time required to complete the task under this constraint.

The coordinate bounds are small, with positions up to 100 and stamina limit up to 100. This immediately tells us that even a brute force over all meaningful states or path patterns would be feasible in principle, but the structure of the problem suggests a direct analytical solution is intended.

A subtle issue is that Monocarp can choose the order of visiting key and chest, and can also optionally relocate the chest while carrying it. That relocation is exactly what interacts with the stamina constraint, and ignoring it leads to incorrect assumptions about always just “visit key then chest” or vice versa.

Approaches

A straightforward way to think about the task is to enumerate all possible strategies: decide whether to pick up the key first or the chest first, decide where to move next, and decide whether to ever carry the chest during movement.

This brute-force viewpoint is correct because any valid solution is a sequence of moves along the line, with occasional pickups and a single final opening action. However, the number of possible strategies grows rapidly once we allow arbitrary intermediate dropping and re-picking of the chest. In the worst case, simulating all walk sequences with state tracking of position, whether the key is held, whether the chest is held, and how much stamina remains leads to an explosion of possibilities that is unnecessary given the problem’s structure.

The key observation is that there are only two meaningful global orders of interaction. Either Monocarp goes to the key first and then to the chest, or he goes to the chest first and then tries to reach the key in a compatible way. Everything else reduces to one of these two templates because pickups are instantaneous and there is no benefit in revisiting the same role more than once.

The only non-trivial decision is what happens when he goes to the chest first: he may choose to carry it while moving toward the key, but that is only beneficial if he is allowed to carry it for long enough. Since carrying cost accumulates only when the chest is in hand, the constraint only affects how much of the segment between x and y can be traversed while carrying it.

This reduces the problem to comparing two candidate plans and validating whether the “carry chest toward key” plan is feasible under k.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over all movement states	Exponential	O(1) or large state space	Too slow
Two-strategy evaluation with constraint check	O(1)	O(1)	Accepted

Algorithm Walkthrough

Compute the absolute distance between key and chest, d = |x - y|. This is the only segment where carrying the chest could matter, because the chest only needs to be moved relative to the key.
Consider the plan where Monocarp visits the key first, then goes to the chest. The total cost is |0 - y| + |y - x|, which simplifies to y + d since positions are positive. In this plan, he never needs to carry the chest while moving, so the stamina constraint is irrelevant.
Consider the plan where Monocarp visits the chest first. From 0 to x costs x. After picking up the chest, he would like to move it closer to y so that meeting the key becomes easier. The best he can do is reduce the distance between chest and key, but doing so requires carrying the chest during movement.
Check whether the chest can be carried for the entire distance between x and y. If k >= d, then it is possible to move the chest directly to the key’s position while carrying it continuously.
If the carry is feasible, the cost of this strategy is |0 - x| + |x - y|, which simplifies to x + d. After arriving at y, Monocarp already has both items at the same point and can open the chest.
Take the minimum of the two strategies.

Why it works

Any valid strategy must end with both the key and chest at the same position. Since there are only two special points in the problem, every optimal walk can be rearranged into one where Monocarp first fully commits to visiting one of the two points before moving toward the other. Intermediate detours do not improve the final alignment cost because movement is linear and pickups are free.

The only non-rearrangeable aspect is carrying the chest while moving, since that introduces the stamina constraint. However, carrying only matters on segments where the chest is actively being relocated relative to the key, and that segment is always exactly the distance between x and y. This collapses all complex movement patterns into a single feasibility check on that segment.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    x, y, k = map(int, input().split())
    
    d = abs(x - y)
    
    # Option 1: go to key first, then chest
    cost_key_first = y + d
    
    # Option 2: go to chest first
    cost_chest_first = x + d if k >= d else float('inf')
    
    print(min(cost_key_first, cost_chest_first))

if __name__ == "__main__":
    solve()

The implementation directly encodes the two candidate strategies. The first term corresponds to visiting the key before any interaction with the chest, which avoids the stamina constraint entirely. The second term corresponds to visiting the chest first and optionally transporting it to the key’s position, which is only valid if the total carried distance does not exceed k.

The only subtle point is the feasibility check k >= abs(x - y). This is the exact condition that allows a continuous carry from chest to key; splitting the movement into multiple carries does not help because the total carried distance still accumulates to the same value.

Worked Examples

Example 1

Input:

x = 5, y = 7, k = 2

We compute d = |5 - 7| = 2.

Step	Position	Action	Cost so far	Carry constraint
1	0 → 7	go to key	7	0
2	7 → 5	go to chest	9	0

This gives cost 7 + 2 = 9.

For chest-first strategy, k = 2 and d = 2, so carrying is allowed.

Step	Position	Action	Cost so far	Carry constraint
1	0 → 5	go to chest	5	0
2	5 → 7	carry chest to key	7	2
3	open	done	7	2

This gives cost 5 + 2 = 7.

The second strategy is better.

Example 2

Input:

x = 10, y = 5, k = 0

We compute d = 5.

Key-first strategy:

Step	Position	Action	Cost so far
1	0 → 5	go to key	5
2	5 → 10	go to chest	10

Chest-first strategy is invalid for carrying because k = 0 < 5.

So the answer is 10.

This example shows that when stamina is zero, the solution degenerates into simply visiting the key first and then the chest without any attempt to relocate it.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Only a constant number of arithmetic operations and comparisons are performed
Space	O(1)	No auxiliary data structures are used

The solution is constant time, which is easily within limits even if the constraints were significantly larger than given.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    x, y, k = map(int, input().split())
    
    d = abs(x - y)
    ans1 = y + d
    ans2 = x + d if k >= d else float('inf')
    return str(min(ans1, ans2))

# provided samples
assert run("5 7 2") == "7"
assert run("10 5 0") == "10"

# minimum-like separation
assert run("1 2 0") == "3"

# chest already close but no stamina
assert run("100 1 0") == "199"

# enough stamina to carry chest fully
assert run("5 10 10") == "15"

# symmetric case
assert run("2 8 3") == "9"

Test input	Expected output	What it validates
5 7 2	7	sample correctness with beneficial carrying
10 5 0	10	no stamina forces key-first strategy
1 2 0	3	smallest non-trivial case
100 1 0	199	large separation, no carry allowed
5 10 10	15	full carry feasibility

Edge Cases

A key edge case is when stamina is zero. In this case, any attempt to move the chest while carrying it is impossible, so the only valid strategy is to avoid moving the chest during any segment where it is held. For input 10 1 0, the algorithm correctly computes d = 9, rejects the chest-first strategy, and returns 1 + 9 = 10, corresponding to visiting the key first.

Another edge case occurs when stamina is large enough to cover the full distance between chest and key. For input 5 10 10, we have d = 5 and k = 10, so the chest can be transported directly to the key. The algorithm selects the chest-first strategy and returns 5 + 5 = 10, matching the optimal continuous carry route.

A final subtle case is when the chest is much closer to the start than the key. For input 2 8 3, both strategies are compared directly: key-first yields 8 + 6 = 14, while chest-first is feasible since k >= 6 is false, so only key-first is allowed, ensuring correctness without needing any additional branching logic.