POJ 3623 Best Cow Line
发布日期:2022-02-05 18:27:27 浏览次数:4 分类:技术文章

本文共 2976 字,大约阅读时间需要 9 分钟。

Best Cow Line, Gold
Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 4024   Accepted: 1487

Description

FJ is about to take his N (1 ≤ N ≤ 30,000) 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

6ACDBCB

Sample Output

ABCBCD

Source

USACO 2007 December Gold



首先想到的是从首尾选择最小的一个 这个是显然的
当首尾相同时 可以向内找到第一个不同的 当前面小于或等于时 先取前面的 否则取后面的

话说我咋写了这么长? Orz六行秒杀的XGH大神
#include
    
     #include
     
      #include
      
       using namespace std;int q[3333],top;int head,tail,s[3333];int a,b,c,m;char chin;int find(int x,int y){
       
while(s[x]==s[y]&&x<=y){x++;y--;}
if(x>y)return 0;
if(s[x]<=s[y])return 0;
return 1;}void print(){
int k=0,a;
for(a=1;a<=top;a++)
{
printf("%c",(char)(q[a]+'A'));
k++;
if(k==80)
{
printf("\n");
k=0;
}
}
if(k!=0)printf("\n");}int main(){
scanf("%d\n",&m);
for(a=1;a<=m;a++)
{
scanf("%c\n",&chin);
s[a]=chin-'A';
}
head=1;tail=m;
while(head<=tail)
{
if(s[head]
{
top++;
q[top]=s[head];
head++;
}
else
if(s[head]>s[tail])
{
top++;
q[top]=s[tail];
tail--;
}
else
{
int direct=find(head,tail);
if(direct==0)
{
top++;
q[top]=s[head];
head++;
}
else
 {
top++;
q[top]=s[tail];
tail--;
}
}
}
print();
return 0;}



转载地址:https://blog.csdn.net/li412302070/article/details/12874553 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!

上一篇:POJ 2668 Game of Lines
下一篇:POJ 3666 Making the Grade

发表评论

最新留言

第一次来,支持一个
[***.191.171.4]2022年08月18日 01时35分14秒