HDU 2609 How many（字符串同构，最小表示法）

How many

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2696    Accepted Submission(s): 1162

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

 

Recommend

yifenfei

题意：问有多少个不同构的字符串 （同构字符串：多次将字符串的第一个元素移到最后所得的多个字符串称同构字符串 Eg："abcd" , "bcda" , "cdab" , "dabc" 为同构字符串，而最小表示法表示的字符串则为上述字符串中字典序排序最小的一个，即 "abcd" 。

#include<bits/stdc++.h>
using namespace std;

const int N = 10000 + 10;
char a[N][101], b[N][202], c[N][101];
int n;
struct A{
    char str[101];
};

int getminsub(char *a)
{
    int i = 0,j = 1,len = strlen(a),k = 0; 
    while(i < len && j < len && k < len) 
    {
        if(k == len) break; 
        if(i == j) j++;
        int ni = i + k, nj = j + k;
        if(ni >= len) ni -= len; 
        if(nj >= len) nj -= len;
        if(a[ni] > a[nj])
        {
            i += k + 1;
            k = 0;
        }
        else if(a[ni] < a[nj])
        {
            j += k + 1;
            k = 0;
        }
        else k++;
    }
    return  i; 
}

int cmp(const void *a,const void *b)
{
    A *c, *d;
    c = (A *)a;
    d = (A *)b;
    return strcmp(c -> str, d -> str);
}

int main()
{
    while(scanf("%d", &n) == 1)
    {
        int l;
        for(int i = 0; i < n; i++)
        {
            scanf("%s", a[i]);
            if(!i) l = strlen(a[i]);
            sprintf(b[i], "%s%s", a[i], a[i]);
            int x = getminsub(a[i]);
            b[i][x + l] = '\0';
            sprintf(c[i], "%s", b[i] + x);
        }
        qsort(c, n, sizeof(c[0]), cmp);  ///这里将字符串按字典序排序
        int sum = 1;
        for(int i = 1; i < n; i++)
        {
            if(strcmp(c[i], c[i - 1]))
                sum++;
        }
        printf("%d\n", sum);
    }
}