紫书第三章训练 UVA 1584 Circular Sequence by 16 BobHuang

Some DNA sequences exist in circular forms as in the following figure, which shows a circular sequence ``CGAGTCAGCT", that is, the last symbol ``T" in ``CGAGTCAGCT" is connected to the first symbol ``C". We always read a circular sequence in the clockwise direction.

Since it is not easy to store a circular sequence in a computer as it is, we decided to store it as a linear sequence. However, there can be many linear sequences that are obtained from a circular sequence by cutting any place of the circular sequence. Hence, we also decided to store the linear sequence that is lexicographically smallest among all linear sequences that can be obtained from a circular sequence.

Your task is to find the lexicographically smallest sequence from a given circular sequence. For the example in the figure, the lexicographically smallest sequence is ``AGCTCGAGTC". If there are two or more linear sequences that are lexicographically smallest, you are to find any one of them (in fact, they are the same).

Input 

The input consists of T test cases. The number of test cases T is given on the first line of the input file. Each test case takes one line containing a circular sequence that is written as an arbitrary linear sequence. Since the circular sequences are DNA sequences, only four symbols, A, C, G and T, are allowed. Each sequence has length at least 2 and at most 100.

Output 

Print exactly one line for each test case. The line is to contain the lexicographically smallest sequence for the test case.

The following shows sample input and output for two test cases.

Sample Input 

2                                     
CGAGTCAGCT                            
CTCC

Sample Output 

AGCTCGAGTC 
CCCT
又是AGCT，又是DNA，这不是运载体什么的环状DNA么，看样例读题意就是找到从环上那个点开始使它的字典序最小，这不是直接套string么，很短很易懂。

#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
    string s;
    cin>>s;
    string mi=s;
    for(int i=1;s[i];i++){
        string c=s.substr(i)+s.substr(0,i);
        if(mi>c)
            mi=c;
    }
    cout<<mi<<endl;
}
}