HDU 2609 How many（最小表示法）

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

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

题目大意：给定n个字符串，字符串可以循环变换，如果变换出来的字符串相同即为同一个字符串，求n个字符串中有几个字符串

题目思路：将所以字符串转换为最小字典序（最小表示法），然后用map标记统计一下个数即可。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
const int N = 1e6+5;

map<string,int>mp;
string str[10005];

int minResprent(string s)
{
    int n = s.length();
    int i=0,j=1,k=0;
    while(i<n&&j<n&&k<n)
    {
        int t = s[(i+k)%n] - s[(j+k)%n];
        if(t==0)
            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()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
            cin>>str[i];
        mp.clear();
        int ans=0;
        for(int i=0;i<n;i++)
        {
            int len = str[i].length();
            int loc=minResprent(str[i]);
            string s = "";
            for(int j=loc;j<len;j++)
                s+=str[i][j];
            for(int j=0;j<loc;j++)
                s+=str[i][j];
            //cout<<"s="<<s<<endl;
            if(mp[s]==0)
            {
                ans++;
                mp[s]=1;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}