HDU 2609 最小表示法

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<vector>
#include<set>
using namespace std;
const int MAXN = 10000 + 5;
typedef long long int LL;
#define INF 0x3f3f3f3f
int MinRepresentation(char *s,int l){
    int i = 0, j = 1, k = 0;
    while (i < l&&j < l&&k < l){
        int li, lj;
        li = (i + k) >= l ? i + k - l : i + k;
        lj = (j + k) >= l ? j + k - l : j + k;
        if (s[li] == s[lj]){
            k++;
        }
        else{
            if (s[li]>s[lj]){
                i = i + k + 1;
            }
            else{
                j = j + k + 1;
            }
            if (i == j){
                j++;
            }
            k = 0;
        }
    }
    return i < j ? i : j;
}
char str[MAXN][105];
set<string>se;
int main()
{
    int n;
    while (~scanf("%d", &n)){
        se.clear();
        for (int i = 0; i < n; i++){
            scanf("%s", str[i]);
            int len = strlen(str[i]);
            int start = MinRepresentation(str[i], len);
            string minstr;
            for (int j = 0; j < len; j++){ //得到字典序最小的同构串
                minstr += str[i][(j + start) % len];
            }
            if (!se.count(minstr)){
                se.insert(minstr);
            }
        }
        printf("%d\n",se.size());
    }
    return 0;
}