POJ 3617 Best Cow Line【贪心】

最新推荐文章于 2021-01-06 16:42:15 发布

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本文介绍了一个有趣的算法问题：如何通过特定的操作将牛只按字母顺序重新排列，以便参加比赛时能够尽早被评审。该问题涉及数据结构和算法设计，通过不断选择原序列的首尾元素并将其添加到新序列中，最终实现最小字典序的目标。

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Description

FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds them past the judges.

The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows’ names.

FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.

FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he’s finished, FJ takes his cows for registration in this new order.

Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.

Input

Line 1: A single integer: N
Lines 2..N+1: Line i+1 contains a single initial (‘A’..’Z’) of the cow in the ith position in the original line

Output

The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows (‘A’..’Z’) in the new line.

Sample Input

6
A
C
D
B
C
B

Sample Output

ABCBCD

Code

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        char data[] = new char[N];

        for (int i = 0; i < N; i++)
            data[i] = sc.next().charAt(0);

        int a = 0, b = N - 1, count = 0;
        boolean flag = true;

        while (a <= b) {
            for (int i = 0; a + i <= b; i++) {
                char head = data[a + i];
                char tail = data[b - i];

                if (head < tail) {
                    flag = true;
                    count++;
                    break;
                } else if (head > tail) {
                    flag = false;
                    count++;
                    break;
                }
            }

            if (flag)
                System.out.printf("%c", data[a++]);
            else
                System.out.printf("%c", data[b--]);

            /* 注意换行 */
            if (count % 80 == 0)
                System.out.println();
        }
        System.out.println();
    }

}