HDOJ 2609 How many

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2609

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题目：

Problem Description

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me
How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some).
For example 0110 express a necklace, you can rotate it. 0110 -> 1100 -> 1001 -> 0011->0110.

 

Input

The input contains multiple test cases.
Each test case include: first one integers n. (2<=n<=10000)
Next n lines follow. Each line has a equal length character string. (string only include '0','1').

 

Output

For each test case output a integer , how many different necklaces.

 

Sample Input

4

0110

1100

1001

0011

4

1010

0101

1000

0001

 

Sample Output

1 2

 

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题意：

多组输入

给出若干字符串 字符串可以通过旋转循环相同 判断有多少不同的字符串

 

思路：

需要理解什么是旋转循环相同

用到字符串中最小表示法的定义 可以求与某个字符串循环同构的字符串中 字典序最小的是哪一个

 

 

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代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn=110;
int n;
string s;
map<string,int>mp;

int getmin(int n){
    int i=0,j=1,k=0,t;
    while(i<n && j<n && k<n){
        t=s[(i+k)%n]-s[(j+k)%n];
        if(!t) k++;
        else{
            if(t>0) i+=k+1;
            else j+=k+1;
            if(i==j) j++;
            k=0;
        }
    }
    return i<j?i:j;
}

int main(){
    while(~scanf("%d",&n)){
        mp.clear();
        int ans=0;
        for(int i=1;i<=n;i++){
            cin>>s;
            int len=s.length();
            int k=getmin(len);
            string ss="";
            for(int j=0;j<len;j++){
                ss+=s[(k+j)%len];
            }
            //cout<<ss<<endl;
            if(mp[ss]==0){
                ans++;
                mp[ss]=1;
            } 
        }
        printf("%d
",ans);
    }
    return 0;
}