HDU——2609How many（字符串的最小表示法+substr）

 

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1852    Accepted Submission(s): 763

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
   
   
   
   

 

题意：输入n串二进制的字符串，将其看成项链，问有多少种项链，第一次WA了发现并不是直接map就可以做的，还要将字符串先转换成一种统一的，比如abcde（为了方便表示就用字母了）与cdeab是一样的，相当于把这串字符串头尾连起来，从不同的视角看过去就显示不一样了，那重点就是把这些不同角度的项链统一成一个角度的。按照最小字典序进行统一。最小表示法的模版是网上找的，但是这个模版只能返回一个开始位置整数，拿来修改了一下就可以了。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<map>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;
string zxc(string s,const int &l)
{
    int i = 0, j = 1, k = 0, t;
    while(i < l && j < l && k < l) 
    {
        t = s[(i + k) >= l ? i + k - l : i + k] - s[(j + k) >= l ? j + k - l : j + k];
        if(!t) k++;
        else
        {
            if(t > 0) i = i + k + 1;
            else j = j + k + 1;
            if(i == j) ++ j;
            k = 0;
        }
    }
    int q=min(i,j);//这里本要返回q（最小表示的起始位置），这里先利用一下
    s+=s;//首先要把字符串自增一下
    s=s.substr(q,l);//（起始位置,原来长度）
    return s;//返回最小表示的字符串
}
int main (void)
{
    string s;
    int t,n;
    while (cin>>n)
    {
        map<string,int>list;//无脑map开始了......
        for (int i=0; i<n; i++)
        {
            cin>>s;
            list[zxc(s,s.size())]=0;
        }
        cout<<list.size()<<endl;
    }
    return 0;
}