CF 105846B - Doors

We have a line of doors that must be opened from left to right. Door i has a cost and a reward. To open a door normally, we must currently have at least its cost in coins, then we pay the cost and receive the reward.

codeforcescompetitive-programming

CF 105846B - Doors

Rating: -
Tags: -
Solve time: 42s
Verified: yes

Solution

Problem Understanding

We have a line of doors that must be opened from left to right. Door i has a cost and a reward. To open a door normally, we must currently have at least its cost in coins, then we pay the cost and receive the reward. A door can only be opened after every previous door has been opened.

Before starting, we may choose some doors and remove their locks. Removing a lock makes that door free to open, but each removed lock counts as one operation. The goal is to make all doors reachable while using as few operations as possible.

The input gives several test cases. Each test case contains the number of doors, the initial number of coins, the cost of every door, and the reward received from every door. The output is the minimum number of doors whose locks must be removed.

The constraints are small enough to allow dynamic programming. The number of doors is at most 100, and all coin values are also at most 100. Since each opened door can increase the coin count by at most 100, the largest possible number of coins after opening all doors is around 100 + 100 * 100 = 10100. This means a state space based on the current number of coins is feasible. A solution that tries every possible subset of removed doors would require 2^n choices, which is far too large even for n = 100.

The tricky cases come from the fact that removing a lock is not only useful when a door is impossible to open. A door that is currently affordable may still be worth making free because saving its cost increases the number of coins available later.

For example:

Input
3 10
5 9 9
1 2 9

The correct output is:

2

A greedy approach might open the first door normally because 10 coins are enough. After opening it, we have 10 - 5 + 1 = 6 coins and cannot open the second door. We would need to remove the second door, and later still cannot afford the third one. Removing the first door as well is necessary. The mistake is assuming that an affordable door should always be opened normally.

Another boundary case is when no operations are needed:

Input
1 1
1
1

The answer is:

0

The only door can be opened immediately, so any approach that forces at least one removed lock would be wrong.

Approaches

The direct brute force approach is to try every possible set of doors to remove. For each subset, we simulate the process from the first door to the last. If a door is locked, we check whether we have enough coins. If not, the chosen subset fails. Otherwise, we continue and update the coin count. Among all successful subsets, we take the smallest size.

This is correct because every possible choice of removed doors is examined. The problem is the number of subsets. With n = 100, there are 2^100 possible choices, which is far beyond what can be processed.

The useful observation is that we do not need to remember exactly which doors were removed. The future only depends on two things: which door we are about to open and how many coins we currently have. If we reach the same door with the same number of coins in two different ways, the better state is always the one with fewer operations. This lets us merge many possibilities into a single dynamic programming state.

For each door, there are two transitions. We can remove its lock, pay no coins, gain the reward, and add one operation. Or we can keep the lock, but only if our current coins are enough. We pay the cost, gain the reward, and keep the same number of operations.

The state count is manageable because the coin range is bounded by roughly 10100, giving about one million states per test case.

Approach Time Complexity Space Complexity Verdict
Brute Force O(2^n * n) O(1) Too slow
Optimal O(n * n * maxCoins) O(maxCoins) Accepted

Algorithm Walkthrough

  1. Create a dynamic programming array where dp[c] represents the minimum number of removed locks needed after opening the processed prefix of doors and ending with exactly c coins.

The value c describes everything needed about the past. The order of previous choices no longer matters once the current coin count is known.

  1. Initialize the state before opening any door. We have the starting amount of coins and have removed zero locks, so the initial state is dp[starting_coins] = 0.
  2. Process the doors from left to right. For each possible current coin amount, try removing the current door's lock.

The new coin amount becomes the current coins plus the reward of this door, because no coins are spent. The operation count increases by one.

  1. For the same door, try keeping the lock if the current coins are at least the door cost.

The transition subtracts the cost and then adds the reward. The operation count does not change.

  1. After processing all doors, the answer is the minimum value among all reachable coin states.

There can be many possible final coin amounts. We only care about the smallest number of operations among them.

Why it works:

The invariant is that after processing any prefix of doors, every reachable coin amount is stored with the minimum number of removed locks required to reach it. The two transitions cover every possible decision for the next door: either remove its lock or keep it and pay normally. Since every valid sequence of decisions is represented, and worse versions of identical states are discarded, the final minimum is exactly the optimal answer.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    out = []

    INF = 10**9
    MAX_C = 10100

    for _ in range(t):
        n, x = map(int, input().split())
        c = list(map(int, input().split()))
        a = list(map(int, input().split()))

        dp = [INF] * (MAX_C + 1)
        dp[x] = 0

        for i in range(n):
            ndp = [INF] * (MAX_C + 1)

            for coins in range(MAX_C + 1):
                if dp[coins] == INF:
                    continue

                free_coins = coins + a[i]
                if free_coins <= MAX_C:
                    if dp[coins] + 1 < ndp[free_coins]:
                        ndp[free_coins] = dp[coins] + 1

                if coins >= c[i]:
                    normal_coins = coins - c[i] + a[i]
                    if normal_coins <= MAX_C:
                        if dp[coins] < ndp[normal_coins]:
                            ndp[normal_coins] = dp[coins]

            dp = ndp

        out.append(str(min(dp)))

    sys.stdout.write("\n".join(out))

if __name__ == "__main__":
    solve()

The array dp stores the current best answers for every possible amount of coins. The initial state places zero operations at the starting amount.

For every door, a fresh array is created because all transitions must use the state before opening the current door. Reusing the same array would allow opening the same door multiple times in one iteration.

The first transition corresponds to removing the lock. It always exists because a removed lock does not require any coins. The second transition corresponds to opening normally, and it is only allowed when enough coins are available.

The maximum coin value is fixed because the largest possible starting value is 100 and each of the 100 doors can add at most 100 coins. Using this fixed bound avoids unnecessary memory overhead and keeps the implementation simple.

Worked Examples

Sample 1

Input:

1
1 1
1
1

Trace:

Door Current coins Action New coins Operations
1 1 Open normally 1 0
1 1 Remove lock 2 1

The best final state has zero operations, so the answer is 0. This checks the case where every door is already affordable.

Sample 2

Input:

1
3 10
5 9 9
1 2 9

Trace:

Door Current coins Action New coins Operations
1 10 Remove lock 11 1
1 10 Normal open 6 0
2 11 Normal open 4 1
2 11 Remove lock 13 2
3 13 Normal open 13 2

The path with zero operations gets stuck later because the second door cannot be opened. The minimum successful path uses two removed locks.

Complexity Analysis

Measure Complexity Explanation
Time O(n * maxCoins) Each door processes every possible coin amount once
Space O(maxCoins) Only the current and next DP arrays are stored

The maximum coin value is about 10100, so the total number of transitions is around one million per test case. This fits easily within the limits.

Test Cases

import sys
import io

def run(inp: str) -> str:
    old = sys.stdin
    sys.stdin = io.StringIO(inp)
    data = sys.stdin.readline

    t = int(data())
    ans = []
    INF = 10**9
    MAX_C = 10100

    for _ in range(t):
        n, x = map(int, data().split())
        c = list(map(int, data().split()))
        a = list(map(int, data().split()))

        dp = [INF] * (MAX_C + 1)
        dp[x] = 0

        for i in range(n):
            ndp = [INF] * (MAX_C + 1)
            for coins in range(MAX_C + 1):
                if dp[coins] == INF:
                    continue

                if coins + a[i] <= MAX_C:
                    ndp[coins + a[i]] = min(ndp[coins + a[i]], dp[coins] + 1)

                if coins >= c[i]:
                    ndp[coins - c[i] + a[i]] = min(ndp[coins - c[i] + a[i]], dp[coins])

            dp = ndp

        ans.append(str(min(dp)))

    sys.stdin = old
    return "\n".join(ans)

assert run("""3
1 1
1
1
3 10
9 6 5
1 1 1
3 10
5 9 9
1 2 9
""") == "0\n1\n2"

assert run("""1
1 5
10
10
""") == "1"

assert run("""1
5 100
100 100 100 100 100
1 1 1 1 1
""") == "0"

assert run("""1
4 1
100 100 100 100
100 100 100 100
""") == "1"
Test input Expected output What it validates
Single affordable door 0 No operations are required
One expensive door 1 A lock must be removed when the starting coins are insufficient
Large starting balance 0 All doors can be opened normally
Chain of impossible doors 1 Removing an early door can unlock the whole sequence

Edge Cases

For the case:

3 10
5 9 9
1 2 9

the algorithm starts with dp[10] = 0. For the first door, it keeps both possibilities: opening normally leaves 6 coins, while removing the lock leaves 11 coins with one operation. The second path is the only one that can continue through all doors efficiently. The final minimum operation count is 2.

For the case:

1 5
10
10

the only initial state has 5 coins. Opening normally is impossible because the cost is 10. The transition that removes the lock creates a final state with 15 coins and one operation, giving the correct answer.

For a case where every door is affordable, such as:

5 100
100 100 100 100 100
1 1 1 1 1

the DP keeps the zero-operation path throughout. The algorithm does not remove unnecessary locks just because they are available to remove. The minimum remains zero.