POJ--3617 Best Cow Line

Best Cow Line

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15789 Accepted: 4465

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 theith 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

字典序最小问题，不断从字符串的开头和末尾选出较小的字符放在新字符串的末尾即可。

以下是我的AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<string>#include<vector>#define rep(i,a,n) for(int i=a;i<n;i++)#define per(i,n,a) for(int i=n;i>a;i--)#define mem0(x) memset(x,0,sizeof(x))#define memff(x) memset(x,INF,sizeof(x))#define INF 100000000using namespace std;char aa[2050];typedef pair<int,int> p;typedef long long LL;int main(){//freopen("input.in","r",stdin);int n;while(cin>>n){for(int i=0;i<n;i++){cin>>aa[i];  } int a=0,b=n-1,cnt=0; while(a<=b) { int flag=0; for(int i=0;i<b;i++) { if(aa[a+i]>aa[b-i]) { flag=0;break; } else if(aa[a+i]<aa[b-i]) { flag=1;break; } } if(flag)cout<<aa[a++]; else cout<<aa[b--]; cnt++; if(cnt%80==0)cout<<endl; } if(n%80)cout<<endl;}return 0; }