POJ 3617 Best Cow Line

最新推荐文章于 2021-02-06 12:21:21 发布

lizhaowei213

最新推荐文章于 2021-02-06 12:21:21 发布

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本文介绍了一种算法，用于将一组字符按照最小字典序排列。通过删除头部或尾部字符并插入到新字符串中，实现从原始排列到最小字典序排列的转换。

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Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15366		Accepted: 4343

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

题目大意，给一个字符串，只能删除头或尾部的字符，并将其插入新字符串尾部，求产生的字典序最小字符串。

方法：贪心，翻转原字符串，和原串从头比较即可。相同位置的元素就比较这一位后的下一个。

#include <stdio.h>
const int MAXN=2005;
char initialString[MAXN],reverseString[MAXN],ansString[MAXN];
int n;

bool isInitialSmall(int ini,int rev)
{
        if(ini==n||rev==n)
                return true;
        if(initialString[ini]<reverseString[rev])
                return true;
        else if(initialString[ini]>reverseString[rev])
                return false;
        else if(initialString[ini]==reverseString[rev])
                return isInitialSmall(ini+1,rev+1);
}

int main()
{
        while(scanf("%d",&n)>0)
        {
                for(int i=0;i<n;i++)
                {
                        char temp[5];
                        scanf("%s",temp);
                        initialString[i]=temp[0];
                        reverseString[n-i-1]=temp[0];
                }
                int initialPos=0,reversePos=0,cnt=0;
                while(cnt!=n)
                {
                        if(initialString[initialPos]<reverseString[reversePos])
                                ansString[cnt++]=initialString[initialPos++];
                        else if(initialString[initialPos]>reverseString[reversePos])
                                ansString[cnt++]=reverseString[reversePos++];
                        else if(initialString[initialPos]==reverseString[reversePos])
                        {
                                if(isInitialSmall(initialPos,reversePos))
                                        ansString[cnt++]=initialString[initialPos++];
                                else
                                        ansString[cnt++]=reverseString[reversePos++];
                        }
                }
                int c=0;
                for(int i=0;i<n;i++)
                {
                        printf("%c",ansString[i]);
                        c++;
                        if(c%80==0&&i<n-1)
                                printf("\n");
                }
        }
        return 0;
}