HDOJ 2609 How many （最小表示法）

How many

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 990 Accepted Submission(s): 382

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

解题思路：

最开始的时候用暴力。即将字符串复制一遍，然后一个个查找。果断TLE...

后来用了kmp优化，还是TLE

查了资料，原来处理字符串的循环同构可以用最小表示法。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char str[110];struct node{    char ch[110];}st[10010];int min_express(){    int len=strlen(str);    int i=0,j=1,k=0;    while(i<len && j<len && k<len)    {        int t;        t=str[(i+k)%len]-str[(j+k)%len];        if (t==0) k++;        else        {            if (t>0) i=i+k+1;            else j=j+k+1;            if (i==j) j++;            k=0;        }    }    return i<j? i:j;}bool cmp(node a, node b){    return (strcmp(a.ch, b.ch)<0);}int main (){    int n,i,j,t=0,len;    while (cin>>n)    {        for (i=0; i<n; i++)        {            t=0;            scanf("%s",str);            len=strlen(str);            int k=min_express();            for (j=k; j<len; j++)                st[i].ch[t++]=str[j];            for (j=0; j<k; j++)                st[i].ch[t++]=str[j];            st[i].ch[t]='\0';        }        sort(st,st+n,cmp);        int sum=0;        for (i=0; i<n-1; i++)            if (strcmp(st[i].ch, st[i+1].ch)!=0)                sum++;        cout<<sum+1<<endl;    }    return 0;}