POJ3617 Best Cow Line Greedy

给你一个字符串，每次两种操作将字符串重组，字典序最小。

将第一个字符加在新字符串尾部

将最后一个字符加在新串尾部

从前从后比较两个字符串字典序，选择小的第一个字符添加到新串尾部，反复如此。相同时随便选一个。不能够简单比较第一个字符。

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5946		Accepted: 1922

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

#include
#include
#include
#include
#include

using namespace std;

#define MAXN 2100

char str[MAXN];
int n;

void func()
{
    int cnt=0,l=0,r=n-1;
    while(l<=r)
    {
        bool lf=1;
        for(int i=0;i+l<=r;i++)
        {
            if(str[l+i]str[r-i])
            {
                lf=0;
                break;
            }
        }

        if(lf)
            putchar(str[l++]);
        else
            putchar(str[r--]);
        cnt++;
        if(cnt==80)
        {
             printf("\n");
             cnt=0;
        }
    }
    printf("\n");
}

int main()
{
    while(cin>>n)
    {
        for(int i=0;i>str[i];
        str[n]='\0';
        func();
    }
    return 0;
}