题目链接：https://ac.nowcoder.com/acm/contest/993/L​​​​​​​/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

题目描述

FJ is about to take his N (1 ≤ N ≤ 500,000) 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.

输入描述

* 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

输出描述

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.

输入

6
A
C
D
B
C
B

输出

ABCBCD

说明

Step Original New
#1 ACDBCB
#2 CDBCB A
#3 CDBC AB
#4 CDB ABC
#5 CD ABCB
#6 D ABCBC
#7 ABCBCD

解题思路

题意:已知一段长度为N的字符串,让你构造一个字典序最小的字符串。
构造的规则如下:如果原始字符串的头部 < 原始字符串的尾部,则从原始字符串的头部删除该字符添加到新的字符串的一个字符;如果头部 > 尾部则删除尾部的字符添加到新字符串中。
思路:从当前字符串最前面，最后面选一个字典序较小的然后拉到一个新的字符串序列中，如果相同就一直往中间扫描直到发现不同为止(一个字符如果被选中之后那么就不可以再次选择了)，所以我们左右各设一个指针扫描就好了，不需要递归。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
char a[30100];
int n, cnt = 0;
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf(" %c", &a[i]);
    int l = 1, r = n;
    while (l < r) {
        if (a[l] < a[r])
            printf("%c", a[l++]);
        else if (a[l] > a[r])
            printf("%c", a[r--]);
        else {
            int x = l, y = r;
            while (x < y && a[x] == a[y])
                x++, y--;
            if (a[x] < a[y])
                printf("%c", a[l++]);
            else printf("%c", a[r--]);
        }
        cnt++;
        if (cnt == 80)
            printf("\n"), cnt = 0;
    }
    printf("%c", a[l]);
    return 0;
}