poj 3617Best Cow Line(贪心)

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11218		Accepted: 3326

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：

给定长度为n 的字符串S,要构造一个长度为n的字符窜T。起初，T 是一个空串，随后反复进行下列操作。

*从S的头部删除一个字符，加到T的尾部

*从S的尾部删除一个字符，加到T的尾部

目标是构造字典序尽可能小的字符串T。

1<=n<=2000

字符串中只包含大写字母。

思路：

*按照字典序比较S和将S反转后的字符串S'

*如果S较小，就从S的开头取一个字母，追加到T的末尾。

*如果S'较小，就从S的末尾取一个字符，追加到T的末尾。

*（如果相同，取哪个都一样）

字典序的比较经常用到贪心。

#include<stdio.h>
#include<string.h>
int n;
char s[2005];
void solve()
{
	int a=0,ans=0;
	int b=n-1;
	while(a<=b)
	{
		int left=0,i;
		for(i=0;a+i<=b;i++)
		{
			if(s[a+i]<s[b-i])//左<右 
			{
				left=1;
				break;
			}
			else if(s[a+i]>s[b-i])//左>右 
			{
				left=0;
				break;
			}
		}
		if(left) printf("%c",s[a++]);//取左侧值 
		else printf("%c",s[b--]);//取右侧值
		ans++;
		if(ans%80==0)
		printf("\n");
	}
	 printf("\n");
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		int i;
		getchar();
		for(i=0;i<n;i++)
		{
			scanf("%c",&s[i]);
			getchar();
		}
		solve();
	}
	return 0;
}