poj3617贪心

最新推荐文章于 2022-04-19 22:54:06 发布

原创最新推荐文章于 2022-04-19 22:54:06 发布 · 286 阅读

0 ·

CC 4.0 BY-SA版权

本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了如何利用这些技术实现视频内容的智能分析与理解。通过详细解析视频分割原理与实践，以及语义识别的应用场景，旨在为读者提供全面的技术指导，促进在智能视频分析领域的应用与创新。

Best Cow Line

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13682		Accepted: 3924

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<stdio.h>
int main()
{
    char c[2006],s[2006],ans[2006];
    int n;
    scanf("%d",&n);
    getchar();
    for(int i=0,j=0;i<n;++i)
    {
        scanf("%s",c);
        s[j]=c[0]; j++;
    }
    int temp=0;
    for(int i=0,j=n-1;i<n,j>=0;temp++)
    {
        if(i==j)
        {
            ans[temp]=s[i]; break;
        }
        if(s[i]<s[j])
        {
            ans[temp]=s[i]; i++;
        }
        else if(s[i]>s[j])
        {
            ans[temp]=s[j]; j--;
        }
        else
        {
            int p=i+1,q=j-1;
            while(s[p]==s[q])
            {
                p++; q--;
            }
            if(s[p]<s[q])
            {
                ans[temp]=s[i]; i++;
            }
            if(s[p]>s[q])
            {
                ans[temp]=s[j]; j--;
            }
        }
    }
    int count=0;
    for(int i=0;i<n;i++)
        {
            printf("%c",ans[i]);
            count++;
            if(count%80==0 && i<n-1)
                printf("\n");
         }
    return 0;
}

/*
6
A
C
D
B
C
B

ABCBCD
*/