挑战2.2 Best Cow Line（POJ 3617）

最新推荐文章于 2021-12-12 11:50:02 发布

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最新推荐文章于 2021-12-12 11:50:02 发布

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分类专栏：挑战程序设计板刷

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本文介绍了一种在农民年度竞赛中，如何通过最优地重新排列牛群顺序来获得最佳评分的方法。

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A - Best Cow Line

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3617

Appoint description:

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

题目大意：

就是说，给你n个字符，然后让你从这n个字符的首部和尾部依次取出来一个字符比较字典序的大小，把字典序小的放到一个新的串的首部，最后输出这个新串。

解题思路：

直接模拟就好了，设置一个l，r,a和b的大小每次是通过l和r来更新的。

代码：

# include<cstdio>
# include<iostream>
# include<algorithm>
# include<cstring>
# include<string>
# include<cmath>
# include<queue>
# include<stack>
# include<set>
# include<map>

using namespace std;

# define inf 999999999
# define MAX 2000+4

 int n;
char s[MAX];

void input()
{
    cin>>n;
    for ( int i = 0;i < n;i++ )
    {
        char ch;
        cin>>ch;
        s[i] = ch;
    }
}

void solve()
{
    int l = 0,r = n-1;
    int cnt = 0;
    while ( l <= r )
    {
        int a = l,b = r;
        while ( a<b&&s[a] == s[b] )
        {
            a++;
            b--;
        }
        
        if ( s[a] > s[b] )
        {
            putchar(s[r--]);
        }
        else
        {
            putchar(s[l++]);
        }
        cnt++;

        if ( cnt%80==0 )
            cout<<endl;
    }
}


int main(void)
{
    input();
    solve();
    cout<<endl;
	return 0;
}