poj3617字典序问题【贪心】

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16664		Accepted: 4692

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

中文意思：

每次都可以从字符串的首部或者尾部提取字母，使得最后的字符串的字典序最小。

!!!好大一个坑…字符输入是一个一个的！！！

可以用cin.getline();但是时间慢，           也可以用 scanf("%s", tmp);s[i] = tmp[0];

参考了网上的思路发现，还可以用scanf(" %c", &ch)；

在格式串中，空格的意思是匹配输入中的所有换行、TAB、空格，所以加上一个空格，就可以屏蔽掉在输入中的所有的不愉快的因素了。

我的代码是按照快速排序的…然而超时，附在最后……

下面这个代码是参考《挑战编程》的思路

#include 
#include 
char s[2002];
int main(){
int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=0;is[b-i]){
					left=false;
					break;
				}
			}
			if(left)  putchar(s[a++]);
			else putchar(s[b--]); 
			ans++;
			if(ans==80){
				printf("\n");
				ans=0;
			} 	
		}	
		  printf("\n");	
	}
	return 0;
}

这个是我的代码，按照快速排序的思想…然而超时了超时了TT…

#include 
#include 
#include 
using namespace std;

int main(){
	int n;
	char x[2002],y[2002];
	scanf("%d",&n);
	for(int i=0;ix[j]||(x[i]==x[j]&&x[i+1]>x[j-1])){
			y[k++]=x[j];
			j--;
		}
	}
	for(int i=0;i