POJ-3617-Best Cow Line(贪心算法+Java)

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11700   Accepted: 3430 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 Source USACO 2007 November Silver 把题目的主要意思解释一下......英语是硬伤啊. 给定长度为N的字符串S,要构造一个长度为N的字符串T。起初，T是一个空串,随后反复进行下列任意操作。 1.从S的头部删除一个字符，加到T的尾部 2.从S的尾部删除一个字符，加到T的尾部 目标是要构造字典序尽可能小的字符串T。 示例: S="CDB" T="ABC" 开头: S="DB" T="ABCC" 末尾: S="CD" T="ABCB" 输出:( 要求输出的字符串要按照80个字符换行.) 限制条件: 1.1<=N<=2000. 2.字符串S只包含大写英文字母. 首先明确一下什么叫字典序: 字典序时指从前到后比较两个字符串大小的方法。首先比较第一个字符, 如果不同则第1个字符较小的字符串更小，如果相同则继续比较第2个字符...... 如此继续，来比较整个字符串的大小。 有了这个我们的思路就可以明确了.从字典序的性质来看,无论T的末尾有多大， 只要前面部分小就可以。所以我们可以试下这个贪心算法。 不断取S的开头和末尾中较小的一个字符放到T的末尾。 这个算法已经接近正确了，只是针对S的开头和结尾的情形没有定义。 在这种情形下，因为我们希望能够尽早使用更小的字符，所以就要比较 下一个字符的大小。下一个字符也有可能相同，因此完善如下： 按照字典序比较S和将S反转后的字符串S'。 如果S较小，就从S的开头取出一个字符，追加到T的末尾。 如果S'较小，就从S'的开头取出一个字符，追加到T的末尾。 （如果相同任选一个都行！） 代码： /* * 字典序时指从前到后比较两个字符串大小的方法。首先比较第一个字符, * 如果不同则第1个字符较小的字符串更小，如果相同则继续比较第2个字符...... * 如此继续，来比较整个字符串的大小。 * * 主要思路:不断取S的开头和末尾中较小的一个字符放到T的末尾。 */ import java.io.*; import java.util.*; public class Main { public static void main(String[] args) { // TODO Auto-generated method stub Scanner input = new Scanner(System.in); int N = input.nextInt(); input.nextLine(); char S[] = new char[2010]; for (int i = 0; i < N; i++) //Java读取单个字符不方便，处理回车符! { String str = input.nextLine(); S[i] = str.charAt(0); } int a = 0, b = N - 1, count = 0; while (a <= b) { boolean left = false; for (int i = 0; a + i < b; i++) //利用布尔变量记录每一次比较结果 { if (S[a + i] < S[b - i]) { left = true; count++; break; } else if (S[a + i] > S[b - i]) { left = false; count++; break; } } if (left) //根据布尔值进行输出! { System.out.print(S[a++]); } else { System.out.print(S[b--]); } if (count % 80 == 0) System.out.println(); } System.out.println(); } }