hdu2609（最小表示法+set)

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1113    Accepted Submission(s): 442

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

4011011001001001141010010110000001

 

Sample Output

12

 

1 题目要求的是给定n个字符串，找出不同的字符串的个数。由于题目说了，字符串可以进行变换，也就是如果两个字符串相同那么它们的最小表示是相同的。
2 只要求出所有字符串的最小表示，然后利用set存储，最后set的元素个数就是最后的ans

#include <iostream>#include <set>#include <cstring>using namespace std;int len,n;int getMin(string &str, int len){    int i=0,j=1,k=0;    while(i<len && j<len && k<len){        int t = str[(i+k)%len]-str[(j+k)%len];        if(!t) k++;        else{            if(t>0) i=i+k+1;            else j=j+k+1;            if(j==i) j++;            k=0;        }    }    return min(i,j);}int main(){    const int MAXN = 210;    string word,tep;    set<string>s;    while(cin>>n){        s.clear();        for(int i=0; i<n; i++){            cin>>word;            len = word.size();            int pos = getMin(word,len);            tep.assign(word,pos,len-pos);            tep.append(word,0,pos);            s.insert(tep);        }        cout<<s.size()<<endl;    }    return 0;}