How many （HDU-2609）（最小表示法）

 

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

题意：给你n串项链（字符串可以左移），判断有多少种不同的项链

思路：最小表示法模板套用

AC代码：

#include <bits/stdc++.h>
const int maxx=100010;
const int inf=0x3f3f3f3f;
using namespace std;
string a[maxx],b[maxx];
set<string>s;
int Get_min(string p)
{
    int plen=p.size();
    int i=0,j=1,k=0,t;
    while(i<plen && j<plen && k<plen)
    {
        t=p[(i+k)%plen]-p[(j+k)%plen];
        if(!t)
            k++;
        else
        {
            if(t>0) i+=k+1;
            else j+=k+1;
            if(i==j) j++;
            k=0;
        }
    }
    return i>j?j:i;
}
int main()
{
    int n;

    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            b[i]=a[i]+a[i];
        }
        for(int i=0;i<n;i++)
        {
            int ans=Get_min(a[i]);
            s.insert(b[i].substr(ans,a[i].size()));
        }
        cout<<s.size()<<endl;
        s.clear();
    }
    return 0;
}