[最小表示法] hdu 2609 How many

题目链接：

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

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1225    Accepted Submission(s): 476

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
  

 

Author

yifenfei

 

Source

奋斗的年代

 

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题目意思：

给n个有0,1组成的串，求可以经过循环旋转，最后不同的串有多少个。

解题思路：

最小表示法。

先求出每个串的字典序最小的串，存入map里，最后输出map大小即可，也可以存入字典树，最后输出叶子个数。

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 220

map<string,int>myp;
char save[Maxn];
int n;

string cal()
{
    int len=strlen(save);

    int i=0,j=1;

    while(i<len&&j<len)
    {
        int k=0;

        while(k<len&&save[(i+k)%len]==save[(j+k)%len])
            k++;
        if(k>=len)
            break;
        if(save[(i+k)%len]>save[(j+k)%len])
        {
            i=i+k+1;
            if(i==j)
                i++;
        }
        else
        {
            j=j+k+1;
            if(i==j)
                j++;
        }
        //printf("i:%d j:%d k:%d\n",i,j,k);
        //system("pause");

    }
    int tt=min(i,j);

    int cnt=0;
    string res;

    while(cnt<len)
    {
        res+=save[(tt+cnt)%len];
        cnt++;
    }
    return res;

}

int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   while(~scanf("%d",&n))
   {
       myp.clear();

       while(n--)
       {
           scanf("%s",save);
           string a=cal();
           //cout<<": "<<a<<endl;
           //system("pause");
           myp[a]=1;
       }
       printf("%d\n",myp.size());
   }
   return 0;
}