POJ-3617-Best Cow Line(贪心算法+Java)
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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
6ACDBCB
Sample Output
ABCBCD
Source
S="CDB"
S="CD"
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。
/* * 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 stubScanner 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();}}
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