POJ3617——Best Cow Line

Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22772   Accepted: 6207 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

题意：

给定一个字符串，将其重新排列，组成一个字典序尽量小的字符串，重新排列的方式为：原字符串的两端字符比较，较小的一个放到新字符串的开头。

解：

贪心解决，两端开始扫描，找寻较小的字符，如果相同则比较下一个字母，当不同时输出较小一侧的字符。

#include 
#include 
#include 
using namespace std;
char str[2005];

int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int count=0;
		memset(str,0,sizeof(str));
		for(int i=0;i>str[i];
		}
		int a,b;
		a=0;b=n-1;
		while(a<=b)
		{
			int flag=0;
			for(int i=0;a+i<=b;i++)
			{
				if(str[a+i]str[b-i])
				{
					flag=0;
					break;
				}
			}
			if(flag)
				printf("%c",str[a++]);
			else
				printf("%c",str[b--]);
			count++;
			if(count%80==0)
			printf("\n");
		}	
	}
	return 0;
}