CF 1829C - Mr. Perfectly Fine

Rating: 800
Tags: bitmasks, greedy, implementation
Solve time: 1m 8s
Verified: yes

Solution

Problem Understanding

Victor has two independent skills he wants to acquire, and each book he can read either teaches him skill 1, skill 2, both, or none. Every book also has a time cost. The task is to choose a set of books whose combined skills cover both required skills while minimizing the total reading time.

Each book is essentially a small “bundle”: it contributes a bitmask over two bits and has a cost. We are allowed to pick any subset of books, but we care only about the union of acquired skills, not the order or repetition. The goal is to reach the full mask 11 with minimum total cost.

The input size is large across test cases, with total number of books up to 200,000. This immediately rules out any solution that tries all subsets or even pairwise combinations for every test case independently. A quadratic or exponential approach will not survive.

A subtle point is that reading multiple books that teach overlapping skills is allowed, but never beneficial if one book strictly dominates another in both skills and cost. Another edge case is when no combination can cover both skills even though each individual skill exists separately.

For example, if we have books:

3 10
4 01

the answer is 7, but if both books only give 00, the answer is -1 even though books exist.

Another pitfall is assuming a single book might suffice; the correct solution must explicitly handle the case where one book already has 11.

Approaches

A direct brute-force approach would try all subsets of books and compute the union of skills for each subset. Each subset check is O(n), and there are 2^n subsets, which is impossible even for n = 30, let alone 2e5.

We can reduce the structure by noticing that there are only four possible skill states: 00, 01, 10, and 11. This collapses the problem into selecting at most two books that combine to form 11, or possibly a single book that already equals 11.

The key observation is that we only ever need:

the cheapest book that provides 10
the cheapest book that provides 01
the cheapest book that provides 11

Any optimal solution must fall into one of two categories. Either we take one 11 book, or we take one 10 book and one 01 book. No other combination improves on this because adding extra books can only increase cost without improving coverage beyond 11.

Thus the problem reduces to maintaining three minimum values while scanning the input once.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force over subsets	O(2^n)	O(n)	Too slow
Track best candidates	O(n)	O(1)	Accepted

Algorithm Walkthrough

We process each test case independently and reduce all books into a few representative candidates.

Initialize three variables: best_01, best_10, and best_11 as infinity. These track the minimum cost among books that provide each skill pattern. We only care about the cheapest representative of each type.
Iterate over all books. For each book, inspect its binary string and classify it into one of the four types.
If the book gives 11, update best_11 with its minimum cost. This is important because a single book can already solve the problem.
If the book gives 01, update best_01.
If the book gives 10, update best_10.
Ignore 00 books because they never contribute to the final skill set.
After processing all books, compute the best possible answer as the minimum of:

the cost of best_11, and the sum best_01 + best_10. 8. If both best_11 is infinite and either best_01 or best_10 is infinite, then it is impossible to form 11, so output -1.

Why it works

The state space is fully captured by the two bits. Any solution that uses more than two books can be compressed: if it uses multiple 10 books, only the cheapest matters; similarly for 01. If it mixes multiple types, the final combination still reduces to either a single 11 or one 10 plus one 01. Therefore, tracking only the minimum cost per type preserves all optimal constructions.

Python Solution

import sys
input = sys.stdin.readline

INF = 10**18

t = int(input())
for _ in range(t):
    n = int(input())

    best_01 = INF
    best_10 = INF
    best_11 = INF

    for _ in range(n):
        parts = input().split()
        cost = int(parts[0])
        s = parts[1]

        if s == "11":
            best_11 = min(best_11, cost)
        elif s == "10":
            best_10 = min(best_10, cost)
        elif s == "01":
            best_01 = min(best_01, cost)

    ans = min(best_11, best_01 + best_10)

    if ans >= INF:
        print(-1)
    else:
        print(ans)

The code is a direct implementation of the three-candidate reduction. Each book is classified in constant time. The final decision compares only two structural possibilities.

A common implementation mistake is forgetting to initialize candidates to a large value, which would incorrectly allow missing categories to contribute zero. Another is treating "00" as relevant, which is unnecessary and can introduce incorrect minima if mishandled.

Worked Examples

Example 1

Input:

Book	Cost	Type	best_10	best_01	best_11
1	2	00	INF	INF	INF
2	3	10	3	INF	INF
3	4	01	3	4	INF
4	4	00	3	4	INF

Final computation:

min(INF, 3 + 4) = 7

This confirms the algorithm correctly combines one 10 and one 01.

Example 2

Input:

Book	Cost	Type	best_10	best_01	best_11
1	5	11	INF	INF	5
2	8	01	INF	8	5
3	7	10	7	8	5

Final computation:

min(5, 7 + 8) = 5

This shows why direct 11 is always a candidate independent of combinations.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Each book is processed once with O(1) classification and updates
Space	O(1)	Only three scalar variables are maintained

The total number of books across test cases is bounded by 2e5, so the linear scan is comfortably within limits. Memory usage remains constant regardless of input size.

Test Cases

import sys, io

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

    INF = 10**18
    t = int(input())
    out = []

    for _ in range(t):
        n = int(input())
        best_01 = INF
        best_10 = INF
        best_11 = INF

        for _ in range(n):
            cost, s = input().split()
            cost = int(cost)

            if s == "11":
                best_11 = min(best_11, cost)
            elif s == "10":
                best_10 = min(best_10, cost)
            elif s == "01":
                best_01 = min(best_01, cost)

        ans = min(best_11, best_01 + best_10)
        out.append(str(ans if ans < INF else -1))

    return "\n".join(out)

# provided sample
assert run("""6
4
2 00
3 10
4 01
4 00
5
3 01
3 01
5 01
2 10
9 10
1
5 11
3
9 11
8 01
7 10
6
4 01
6 01
7 01
8 00
9 01
1 00
4
8 00
9 10
9 11
8 11
""") == """7
5
5
9
-1
8"""

# single book direct solve
assert run("""1
1
5 11
""") == "5"

# impossible case
assert run("""1
2
3 10
4 10
""") == "-1"

# must combine
assert run("""1
2
3 10
4 01
""") == "7"

Test input	Expected output	What it validates
single 11 book	5	direct solution
only one skill type	-1	impossibility detection
10 + 01 combo	7	correct pairing logic

Edge Cases

When all books are 00, the algorithm keeps all candidates at infinity and correctly outputs -1 because no combination improves the state.

When only one of 01 or 10 exists, the sum becomes impossible since one term remains infinity, preventing a false valid answer.

When multiple optimal candidates exist, the algorithm always retains the minimum cost due to continuous min updates, ensuring correctness even under heavy duplication or reordering.