Best Cow Line POJ - 3617 (贪心字典序最小问题)_best cow line poj

本文介绍了一种在年度“年度农民”比赛中，通过重新排列牛群以实现最小子典序注册字符串的策略。FJ需要将他的牛群按一定顺序排列并注册，为了尽早完成比赛返回农场，他决定调整牛群的顺序，每次从队伍首尾选择一头牛形成新的队伍，以达到最小字典序的目标。

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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

在这里插入图片描述

题解
每次都从字符串的首部或者尾部取一个字母，使最后的字符串的字典序为最小。

注意输入是一个字符一个字符输入！！
使用getchar()运行时间慢容易超时，输入可以用字符输入，注意一下换行，可以用scanf("\n%c",&s[i]);

下面看下代码吧！

#include<stdio.h>
char s[2010];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int i,head,tail;
        for(i=0; i<n; i++)
        {
            scanf("\n%c",&s[i]);    
        }
        head=0;
        tail=n-1;
        int flag,s1=0;
        while(head<=tail)
        {
            flag=0;
            for(i=0; head+i<=tail; i++)
            {        //第一个字符与最后一个字符比大小，依次对称比较
                if(s[head+i]<s[tail-i])  
                {
                    flag=0;
                    break;
                }
                else if(s[head+i]>s[tail-i])
                {
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                printf("%c",s[head++]);
            else
                printf("%c",s[tail--]);
            s1++;
            if(s1==80)
            {
                printf("\n");
                s1=0;
            }
        }
        printf("\n");
    }
    return 0;
}