//从初始情况搜索两层并记录每次状态，这两层中如果有解，直接输出 //若是无解，从有序状态往回搜一层和两层， //如果能达到与之前的相同的状态就是有解。 #include #include #include #include #include #include #include #include #include #include #include #include #define pi acos(-1.0) #define inf 1<<29 #define INF 0x3f3f3f3f #define zero 1e-8 using namespace std; int step, n; int arr[100]; int fin[100]; set ff; int makenum(int num, char *s, int &k) { int i = 0, now[5] = {}; for (; num;) { now[++i] = num % 10; num /= 10; } for (int j = i; j; j--) { s[k++] = now[j] + '0'; } return k; } string makestr(int *s, int num) //样式 { string aa; char a[100]; int k = 0; for (int i = 0; i < num; ++i) { makenum(s[i], a, k); a[k++] = ' '; } a[k] = '\0'; return a; } bool compare(int *s) { for (int i = 0; i < n; ++i) if (fin[i] != s[i]) return false; return true; } void show(int *s) { for (int i = 0; i < n; ++i) cout << s[i]; cout << endl; } void cop(int *en, int *be) { for (int i = 0; i < n; ++i) en[i] = be[i]; } void dfs1(int *tem, int cnt, int now) { string str = makestr(tem, n); ff.insert(str); if (step != inf) return ; if (cnt == now) { if (compare(tem)) { step = cnt; } return ; } if (cnt < now) return ; int tt[100]; for (int j = 0; j < n; ++j) { int z; for (z = 0; z < j; ++z) tt[z] = tem[z]; for (int len = 1; len < n - j; ++len) { cop(tt, tem); int f = z + len; for (int sert = 0; sert < n - j - len; ++sert) { tt[z + sert] = tem[f + sert]; if (f + sert + 1 < n - j - len && tem[f + sert + 1] == tt[z + sert] + 1) continue; for (int kk = 1; kk <= len; ++kk) tt[z + sert + kk] = tem[j + kk - 1]; string str = makestr(tt, n); ff.insert(str); dfs1(tt, cnt, now + 1); } } } } void dfs2(int *tem, int cnt, int now) { if (step != inf) return ; string str = makestr(tem, n); if (cnt == now) { if (ff.find(str) != ff.end()) step = 0; return ; } int tt[100]; for (int j = 0; j < n; ++j) { int z; for (z = 0; z < j; ++z) tt[z] = tem[z]; for (int len = 1; len < n - j; ++len) { cop(tt, tem); int f = z + len; for (int sert = 0; sert < n - j - len; ++sert) { tt[z + sert] = tem[f + sert]; if (f + sert + 1 < n - j - len && tem[f + sert + 1] == tt[z + sert] + 1) continue; for (int kk = 1; kk <= len; ++kk) tt[z + sert + kk] = tem[j + kk - 1]; dfs2(tt, cnt, now + 1); } } } } void makefin() { for (int i = 0; i < n; ++i) fin[i] = i + 1; } int main() { int t; cin >> t; for (; t--;) { scanf("%d", &n); for (int i = 0; i < n; ++i) { scanf("%d", &arr[i]); } makefin(); step = inf; for (int i = 0; i <= 2; ++i) { step = inf; ff.clear(); dfs1(arr, i, 0); if (step != inf) { printf("%d\n", step); break; } } if (step == inf) { for (int i = 1; i <= 2; ++i) { dfs2(fin, i, 0); if (step != inf) { printf("%d\n", i + 2); break; } } if (step == inf) printf("5 or more\n"); } } return 0; }
