CF 1800C1 - Powering the Hero (easy version)

Rating: 1000
Tags: data structures, greedy
Solve time: 3m 50s
Verified: yes

Solution

Problem Understanding

We process a deck of cards from top to bottom. Positive values represent bonus cards, while 0 represents a hero card.

Whenever we encounter a bonus card, we may keep it for later or throw it away. Whenever we encounter a hero card, we may give that hero the strongest available stored bonus. That bonus is consumed and added to the total army power. If no bonus is available, the hero contributes nothing.

The goal is to maximize the total power gained by all heroes.

The input contains multiple test cases. For each test case we are given the deck order and must compute the largest possible total power.

The constraints are small in this easy version. The total number of cards across all test cases is at most 5000. Even an $O(n^2)$ solution would fit comfortably, but there is a much cleaner greedy solution that runs in $O(n \log n)$.

The subtle part of the problem is that bonuses can only be used by heroes appearing after those bonuses in the deck. We cannot arbitrarily match heroes and bonuses.

Consider the deck:

0 100

The answer is 0, not 100. The hero appears before the bonus, so the bonus cannot be used.

Another important case is:

5 1 0

The answer is 5, not 6. A hero can receive only one bonus card. Taking both bonuses is impossible.

A third case is:

1 2 3 0 0

The answer is 5. The first hero should receive 3, the second should receive 2. Giving away 1 first would waste value and produce only 4.

These examples show that whenever a hero appears, we should use the largest available bonus.

Approaches

A brute-force solution would try every possible decision of whether to keep or discard each bonus card and every possible assignment of stored bonuses to heroes. This explores exponentially many states because each bonus introduces multiple choices. Even for a few dozen cards this becomes infeasible.

The key observation is that keeping a bonus card never hurts. If a bonus is not useful later, we can simply ignore it. The only meaningful decision occurs when a hero appears.

Suppose several bonus cards are currently available. If we give the hero anything except the largest bonus, we can swap that choice with the largest one and never decrease the final answer. The current hero gains more power, while the smaller bonus remains available for future heroes.

This exchange argument leads directly to a greedy strategy. Maintain all available bonuses in a max-heap. When a hero appears, remove the largest bonus from the heap and add it to the answer.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	Exponential	Exponential	Too slow
Optimal Heap Greedy	$O(n \log n)$	$O(n)$	Accepted

Algorithm Walkthrough

Initialize an empty max-heap and set the answer to zero.
Process the deck from left to right.
If the current card is a positive number, insert its value into the heap because this bonus is now available for future heroes.
If the current card is a hero card (0), check whether the heap is non-empty.
If at least one bonus is available, extract the largest bonus from the heap and add it to the answer.
Continue until all cards have been processed.
Output the accumulated answer.

Why it works

At every hero card, only bonuses seen earlier can be used. Among those bonuses, choosing the largest available value is always optimal.

Assume a hero receives bonus $x$ while a larger available bonus $y$ exists. Swapping the assignments so that the current hero receives $y$ increases the current contribution by $y-x$. The smaller bonus $x$ remains available for later heroes, so no future choice becomes worse. Repeating this argument eliminates every non-greedy choice and transforms any optimal solution into one that always takes the largest available bonus.

Since the heap always stores exactly the bonuses that can still be assigned, extracting the maximum at each hero card produces an optimal result.

Python Solution

import sys
import heapq

input = sys.stdin.readline

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

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

        heap = []
        total = 0

        for x in cards:
            if x > 0:
                heapq.heappush(heap, -x)  # max-heap
            else:
                if heap:
                    total += -heapq.heappop(heap)

        answers.append(str(total))

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

if __name__ == "__main__":
    solve()

The heap stores negative values because Python's heapq implements a min-heap. Pushing -x allows the smallest stored value to correspond to the largest original bonus.

Every positive card is inserted immediately when it becomes available. Every hero card consumes at most one bonus, matching the rules exactly.

The check if heap: is important because heroes may appear before any bonus card. In that situation the hero contributes zero power.

The answer can become large because card powers reach $10^9$, so we accumulate the result using Python integers.

Worked Examples

Example 1

Input deck:

3 3 3 0 0

Card	Heap Before	Action	Heap After	Answer
3	{}	add bonus	{3}	0
3	{3}	add bonus	{3,3}	0
3	{3,3}	add bonus	{3,3,3}	0
0	{3,3,3}	take 3	{3,3}	3
0	{3,3}	take 3	{3}	6

Final answer: 6.

The trace shows that unused bonuses may remain in the heap after all heroes have been processed.

Example 2

Input deck:

1 2 5 0 4 3 0

Card	Heap Before	Action	Heap After	Answer
1	{}	add	{1}	0
2	{1}	add	{2,1}	0
5	{2,1}	add	{5,2,1}	0
0	{5,2,1}	take 5	{2,1}	5
4	{2,1}	add	{4,2,1}	5
3	{4,2,1}	add	{4,3,2,1}	5
0	{4,3,2,1}	take 4	{3,2,1}	9

Final answer: 9.

This example demonstrates why taking the largest available bonus is essential.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n \log n)$	Each card causes at most one heap insertion or removal
Space	$O(n)$	In the worst case all bonuses are stored in the heap

With at most 5000 cards across all test cases, $O(n \log n)$ runs comfortably within the limits.

Test Cases

import sys
import io
import heapq

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

    input = sys.stdin.readline
    t = int(input())
    out = []

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

        heap = []
        ans = 0

        for x in a:
            if x > 0:
                heapq.heappush(heap, -x)
            elif heap:
                ans += -heapq.heappop(heap)

        out.append(str(ans))

    return "\n".join(out)

# provided sample
assert run(
"""5
5
3 3 3 0 0
6
0 3 3 0 0 3
7
1 2 3 0 4 5 0
7
1 2 5 0 4 3 0
5
3 1 0 0 4
"""
) == "6\n6\n8\n9\n4"

# minimum size
assert run(
"""1
1
0
"""
) == "0"

# hero before bonus
assert run(
"""1
2
0 100
"""
) == "0"

# single hero gets largest bonus
assert run(
"""1
3
5 1 0
"""
) == "5"

# multiple heroes
assert run(
"""1
5
1 2 3 0 0
"""
) == "5"

Test input	Expected output	What it validates
`0`	`0`	Smallest possible deck
`0 100`	`0`	Future bonuses cannot be used
`5 1 0`	`5`	One hero receives only one bonus
`1 2 3 0 0`	`5`	Greedy selection of largest bonuses

Edge Cases

Consider the deck:

0 7

The hero appears before any bonus card. The heap is empty when the hero is processed, so nothing is added to the answer. The later bonus cannot travel backward in time. The algorithm correctly returns 0.

Consider:

10 1 0

When the hero appears, both bonuses are available. The heap contains {10,1} and the algorithm removes 10. The answer becomes 10. Taking 1 instead would permanently lose nine units of power.

Consider:

5 4 3 0 0 0

The first hero receives 5, the second receives 4, and the third receives 3. The answer becomes 12. The heap naturally matches the strongest remaining bonus to each successive hero, which is exactly the optimal assignment.