若两个字符串的长度均为n,则最坏情况下，此程序的时间复杂度为 0（n!）。（）

#include <iostream>
#include <queue>
using namespace std;

const int maxi = 2000000000;

class Map {
    struct item {
        string key; int value;
    } d[maxi];
    int ent;
public:
    int find(string x) {
        for (int i = 0; i < ent; ++i)
            if (d[i].key == x)
                return d[i].value;
        return -1;
    }
    static int end() { return -1; }
    void insert(string k, int v) {
        d[ent].key = k; d[ent++].value = v;
    } S[2];
};

class Queue {
    string q[maxi];
    int head, tail;
public:
    void pop() { ++head; }
    string front() { return q[head + 1]; }
    bool empty() { return head == tail; }
    void push(string x) { q[++tail] = x; }
} q[2];

string st0, st1;
int m;

string LtoR(string s, int L, int R) {
    string t = s;
    char tmp = t[L];
    for (int i = L; i < R; ++i)
        t[i] = t[i + 1];
    t[R] = tmp;
    return t;
}

string RtoL(string s, int L, int R) {
    string t = s;
    char tmp = t[R];
    for (int i = R; i > L; --i)
        t[i] = t[i - 1];
    t[L] = tmp;
    return t;
}

bool check(string st, int p, int step) {
    if (s[p].find(st) != s[p].end())
        return false;
    ++step;
    if (s[p ^ 1].find(st) == s[p].end()) {
        s[p].insert(st, step);
        q[p].push(st);
        return false;
    }
    cout << s[p ^ 1].find(st) + step << endl;
    return true;
}

int main() {
    cin >> st0 >> st1;
    int len = st0.length();
    if (len != st1.length()) {
        cout << -1 << endl;
        return 0;
    }
    if (st0 == st1) {
        cout << 0 << endl;
        return 0;
    }
    cin >> m;
    s[0].insert(st0, 0); s[1].insert(st1, 0);
    q[0].push(st0); q[1].push(st1);
    for (int p = 0; !q[0].empty() && q[1].empty();)
        P ^= 1) (
            string st = q[p].front(); q[p].pop();
            int step = s[p].find(st);
            if ((p == 0 &&
                (check(LtoR(st, 0, m), 90check(RtoL(st, m, en - 1), p, step) ||
                check(RtoL(st, 0, m), p, step)))
                ||
                (p == 1 &&
                (check(LtoR(st, 0, m, p, step) ||
                check(RtoL(st, m, len - 1), p, step))))
                return 0;
        cout << -1 << endl;
        return 0;
    }
}