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USACO 2021 US Open, Silver Problem 3. Acowdemia

Category: 国际竞赛, 计算机国际竞赛 Date: 2022年7月5日上午11:55

USACO 2021 US Open, Silver Problem 3. Acowdemia

Bessie the cow has enrolled in a computer science PhD program, driven by her love of computer science and also the allure of one day becoming "Dr. Bessie". Having worked for some time on her academic research, she has now published? $N$ ?papers ( $1 \leq N \leq 10^{5}$ ), and her? $i$ -th paper has accumulated? $c_{i}$ ?citations ( $0 \leq c_{i} \leq 10^{5}$ ) from other papers in the research literature.

Bessie has heard that an academic's success can be measured by their? $h$ -index. The? $h$ -index is the largest number? $h$ ?such that the researcher has at least? $h$ ?papers each with at least? $h$ ?citations. For example, a researcher with? $4$ ?papers and respective citation counts? $(1, 100, 2, 3)$ ?has an? $h$ -index of? $2$ , whereas if the citation counts were? $(1, 100, 3, 3)$ ?then the? $h$ -index would be? $3$ .

To up her? $h$ -index, Bessie is planning to write up to? $K$ ?survey articles ( $0 \leq K \leq 10^{5}$ ), each citing many of her past papers. However, due to page limits, she can only cite at most? $L$ ?papers in each survey ( $0 \leq L \leq 10^{5}$ ). Of course, no paper may be cited multiple times in a single survey (but a paper may be cited in several surveys).

Help Bessie determine the maximum? $h$ -index she may achieve after writing these survey articles. Bessie is not allowed to cite a survey from one of her surveys.

Note that Bessie's research advisor should probably inform her at some point that writing a survey solely to increase one's? $h$ ?index is ethically dubious; other academics are not recommended to follow Bessie's example here.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains? $N$ ,? $K$ , and? $L$ .The second line contains? $N$ ?space-separated integers? $c_{1}, \dots, c_{N}$ .

OUTPUT FORMAT (print output to the terminal / stdout):

The maximum? $h$ -index on a single line.

SAMPLE INPUT:

4 4 1
1 100 1 1

SAMPLE OUTPUT:

In this example, Bessie may write up to? $4$ ?survey articles, each citing at most? $1$ ?paper. If she cites each of her first and third articles twice, then her? $h$ -index becomes? $3$ .

SAMPLE INPUT:

4 1 4
1 100 1 1

SAMPLE OUTPUT:

In this second example, Bessie may write at most a single article. If Bessie cites any of her first, third, or fourth papers at least once, her? $h$ -index becomes? $2$ .

SCORING:

Test cases 1-6 satisfy? $N \leq 100$ .
Test cases 7-16 satisfy no additional constraints.

Problem credits: Dhruv Rohatgi

USACO 2021 US Open, Silver Problem 3. Acowdemia 题解(翰林国际教育提供，仅供参考)

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(Analysis by Dhruv Rohatgi )

Sort the citation counts largest-to-smallest, and consider a bar graph where bar? $i$ ?has width? $1$ ?and height? $c_{i}$ . The? $h$ -index of the papers before the surveys is simply the dimension of the largest square that fits under this bar graph. We want to determine how much we can increase the? $h$ -index with? $K$ ?surveys each citing? $L$ ?distinct papers.

Let's binary search on the final? $h$ -index. Then we just need a way to check whether it's possible to achieve a given? $h$ -index? $h$ .

It's clearly optimal to work only with the? $h$ ?papers that start off with the largest citation counts. Note that we have? $K L$ ?total citations to allocate to these papers. If the total citation count of these papers is less than? $h^{2} - K L$ , then we cannot hope to achieve? $h$ . This is one failure mode.

Unfortunately, it's not the only failure mode. That is, the converse of the above statement is not true, because we have an added restriction that no survey can cite a paper twice. So for example if? $h = 3$ ,? $K = 1$ , and? $L = 2$ , and the top three papers initially have citation counts? $(3, 3, 1)$ , then we cannot raise the third paper to citation count? $3$ ?since we only have one survey. This illuminates another possible failure mode: if there is a paper with less than? $h - K$ ?citations initially, then we cannot achieve? $h$ .

It turns out that these are the only two failure modes. We can prove this by induction on? $K$ . If? $K = 0$ ?then every paper already has at least? $h$ ?citations, so we're certainly happy. Suppose? $K > 0$ . Every paper needs at most? $K$ ?more citations. There is some set of papers which need exactly? $K$ ?more citations. By the assumption on the sum of citation counts, this set has size at most? $L$ , so with one survey we can cite all these papers (plus some others, until we've hit? $L$ ?citations or this survey has cited every paper). Now we're in a situation with? $K - 1$ ?remaining surveys, but every paper needs at most? $K - 1$ ?more citations. Moreover, it can be checked that the total number of needed citations is now at most? $(K - 1) L$ . So we can indeed induct, completing the proof.

Below is Danny Mittal's code, implementing the above idea.

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Comparator;
import java.util.StringTokenizer;
 
public class AcowdemiaSilver {
 
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer tokenizer = new StringTokenizer(in.readLine());
        int n = Integer.parseInt(tokenizer.nextToken());
        int k = Integer.parseInt(tokenizer.nextToken());
        int l = Integer.parseInt(tokenizer.nextToken());
        Integer[] publications = new Integer[n];
        tokenizer = new StringTokenizer(in.readLine());
        for (int j = 0; j < n; j++) {
            publications[j] = Integer.parseInt(tokenizer.nextToken());
        }
        Arrays.sort(publications, Comparator.reverseOrder());
        int upper = n;
        int lower = 0;
        while (upper > lower) {
            int mid = (upper + lower + 1) / 2;
            long needed = 0;
            for (int j = 0; j < mid; j++) {
                needed += (long) Math.max(0, mid - publications[j]);
            }
            if (needed <= ((long) k) * ((long) l) && publications[mid - 1] + k >= mid) {
                lower = mid;
            } else {
                upper = mid - 1;
            }
        }
        System.out.println(upper);
    }
}

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