Best Cow Line（POJ3617）

最新推荐文章于 2020-09-17 19:15:15 发布

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最新推荐文章于 2020-09-17 19:15:15 发布

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原文链接：http://www.cnblogs.com/kangdong/p/8733440.html

本文介绍了一种通过比较牛群初始字母进行排序的算法。该算法用于决定农民参加年度农民竞赛时牛群展示的顺序。通过将牛群按最小字典序重新排列，以确保农民能尽快完成竞赛并返回农场。文章提供了具体的实现代码。

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
//刚开始没看到要80个字母换行一次，导致一直是格式错误，也是醉了，大佬的又一次复制；

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 using namespace std;
 5 int main()
 6 {
 7     char aa[5000];
 8     int n;
 9     cin >> n;
10     for (int i = 0; i < n; i++)
11         cin >> aa[i];
12     int a = 0, b = n - 1,mark =0;
13     while (a <= b)
14     {
15         bool left = false;
16         for (int i = 0; a + i <= b; i++)
17         {
18             if (aa[a + i] < aa[b - i])
19             {
20                 left = true;
21                 mark++;
22                 break;
23             }
24             else if (aa[a + i] > aa[b - i])
25             {
26                 left = false;
27                 mark++;
28                 break;
29             }
30         }
31         if (left) cout << aa[a++];
32         else cout << aa[b--];
33         if (mark % 80 == 0)cout << endl;
34     }
35     putchar('\n');
36     return 0;
37 }