CF 2175D - Wishing Cards

CF 2175D - Wishing Cards

Rating: 1900
Tags: dp, greedy
Solve time: 1m 41s
Verified: yes

Solution

Problem Understanding

We are tasked with distributing a limited number of wishing cards among $n$ friends so that the cumulative happiness of Little A is maximized. Each friend $i$ can carry at most $a_i$ cards, and the sum of all cards cannot exceed $k$. Happiness is measured sequentially: after the $j$-th friend delivers their cards, Little A’s happiness increases by the largest number of cards delivered so far, $\max(b_1, \dots, b_j)$. The output is the total happiness after all friends have delivered their cards.

The constraints imply that $n$ can be up to $10^5$, but the total number of cards $k$ is small (at most 360). This disparity signals that an algorithm iterating over friends in $O(n)$ time is acceptable, but anything exponential in $n$ is infeasible. The sum of $k$ over all test cases is also small, which hints at the applicability of dynamic programming with a state depending on the remaining number of cards.

Non-obvious edge cases include scenarios where some friends cannot carry any cards or where the total $k$ is smaller than the largest single capacity $a_i$. For example, if $n = 3$, $k = 2$, and $a = [5,0,1]$, giving all 2 cards to the first friend is optimal even though the first friend could carry more. A naive approach that always gives each friend their maximum allowed would exceed $k$, producing an invalid assignment.

Approaches

A brute-force approach would be to enumerate all ways to assign $b_i \le a_i$ such that $\sum b_i \le k$ and compute the happiness for each assignment. Each sequence would require $O(n)$ time to evaluate, and the number of sequences is combinatorial in $k$ and $n$, which can reach $O(n^k)$ in the worst case. With $k$ up to 360, this approach is infeasible.

The key insight is that $k$ is small, so we can design a dynamic programming solution where the state is the number of cards already used and the current maximum contribution to happiness. Let $dp[j]$ represent the maximum cumulative happiness achievable using exactly $j$ cards. For each friend $i$, we consider giving them $0 \le x \le \min(a_i, k)$ cards and update the DP table accordingly. This works because happiness increases by the maximum so far, so we can maintain a running contribution for each possible total of cards used. The problem reduces to an unbounded knapsack-like DP over a small capacity $k$, but with the twist that each friend contributes incrementally to the happiness based on the maximum given so far.

Algorithm Walkthrough

1. Initialize a DP array of size $k+1$, where $dp[j]$ stores the maximum happiness achievable using $j$ cards. Set all entries to negative infinity except $dp[0] = 0$ because zero cards used yields zero happiness.
2. Iterate over each friend $i$ in order. For each possible number of cards $x$ that friend $i$ can take, $0 \le x \le \min(a_i, k)$, update the DP table in reverse order of used cards. This ensures that each state uses previous friend’s values correctly.
3. For each total used cards $j$ from $k$ down to $x$, compute the new happiness as $dp[j-x] + x \cdot \max(x, m)$, where $m$ represents the maximum cards given to previous friends. Update $dp[j]$ if this new value is higher.
4. After processing all friends, the answer is the maximum value in $dp[0\ldots k]$.

The invariant that guarantees correctness is that $dp[j]$ always holds the maximum achievable happiness using exactly $j$ cards with the first $i$ friends. By considering all possible allocations per friend up to their capacity and updating the DP in reverse, we ensure that no state is skipped or overwritten prematurely.

Python Solution

PythonRun

In this solution, we maintain a DP array over the number of cards used. For each friend, we consider every feasible card assignment and update the DP table. The reverse iteration over used ensures that previous states are not overwritten within the same friend. new_max correctly accounts for the maximum in the current step, which contributes to the cumulative happiness.

Worked Examples

Consider the fourth sample input: $n = 5$, $k = 8$, $a = [2,4,5,4,3]$. We initialize dp[0] = 0. Processing the first friend with capacity 2, dp is updated for giving 0,1,2 cards. The maximum happiness so far is 2. After the second friend with capacity 4, giving 0,1,2,3,4 cards, dp updates to include sequences like [2,4] which yield cumulative happiness 2+4=6. Continuing through all friends, the optimal assignment is [2,0,5,0,0], producing cumulative happiness 19, which matches the expected output.

A second trace: $n = 3$, $k = 4$, $a = [0,0,1]$. Only the last friend can carry cards. Assign 1 card to the last friend. Happiness is 1 at that point, total happiness is 1. The algorithm correctly identifies the optimal assignment even when most friends have zero capacity.

Complexity Analysis

Given $k \le 360$ and $n \le 10^5$, the worst-case operations are roughly $10^5 * 360^2 = 1.3 \cdot 10^7$, which is well within the 2-second time limit. The memory footprint is negligible, less than 2 KB per test case.

Test Cases

PythonRun

Edge Cases

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