POJ-3617 Best Cow Line

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

做题学英文

1. phase ：
n.相，周相; 阶段; [物理学]相位; 方面，侧面;
vt.分阶段实行; 调整相位;
vi.分阶段进行;

2. lexicographic
adj.词典的，词典编辑的;

3. marshal
n.元帅; 典礼官; 执法官; 消防局长;
vt.整理，排列，集结;
vi.排列; 编队;

思路：很简单的一道题，假使我们将原始的字符串为S，我们不断寻找S的首尾中最小的元素添加到空字符串T，若首尾相同添加哪个呢？这时候我们比较首尾的下一个元素，添加小的一边首或尾元素，这样依次下去。

#include<iostream>using namespace std;int main(){    int N,cnt=0;    char S[2005];    scanf("%d",&N);    for(int i=0;i<N;i++)    scanf("%s",&S[i]);     int a=0,b=N-1;    while(a<=b){        bool left=false;        for(int i=0;i<=(b-a)/2;i++){     //相同则比较下一对             if(S[a+i]<S[b-i]){  //左边小于右边，从左边取                 left=true;                break;            }            else if(S[a+i]>S[b-i]){     //右边小于左边，从右边取                 left=false;                break;            }        }        if(left) putchar(S[a++]);        else putchar(S[b--]);        cnt++;              //计算字符个数         if(cnt==80)          {              putchar('\n');            cnt=0;          }      }    putchar('\n');    return 0;} 

题目链接：

http://poj.org/problem?id=3617