POJ3617 Best Cow Line【水题】

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27635		Accepted: 7438

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

问题链接：POJ3617 Best Cow Line。

问题简述：

  输入一个正整数N，再输入N行，每行包含一个字母（'A'-'Z'），将这些字母按顺序构成一个字符串S。按照以下规则取字符，顺序构成一个新的字符串T，T是按照字典顺序最小的。

  1.从S的头取一个字符并且删除该字符，连接到T中（开始时T为空串）；

  2.从S的尾取一个字符并且删除该字符，连接到T中。

问题分析：

  这是用一个字符串构建另外一个字符串的问题。比较S的首字符和尾字符，取其小连接到T中即可；当首尾字符相同时，则需要比较下一个，尽量取下一个较小的；当S是一个回文串时，则从哪边取字符结果都是一样的。

程序说明：

  使用一个函数test()来比较哪边更小，是一个最为有效的方法，可以使得程序逻辑更加简洁。需要注意的一点是，输出时，每行最多输出80个字符，即每80个字符输出一个换行。

  比起使用排序的程序，这个程序要快速一些。

AC的C++语言如下：

/* POJ3617 Best Cow Line */

#include 

using namespace std;

//#define DEBUG

const int LEFT = 1;
const int RIGHT = 2;

const int MAXN = 2000;
char a[MAXN+1];

int test(char s[], int start, int end)
{
    while(start < end) {
        if(s[start] < s[end])
            return LEFT;
        else if(s[start] > s[end])
            return RIGHT;

        start++;
        end--;
    }

    return LEFT;
}

int main()
{
    int n;

    cin >> n;
    for(int i=0; i> a[i];
    a[n] = '\0';

#ifdef DEBUG
    cout << a << endl;
#endif

    int start=0, end=n-1, count=0;
    for(int i=0; i