Ultraviolet A 141-the spot game

In fact, it is required to compare four graphs (three of which are obtained by rotation flip) each time. Then, if none of the four figures have been shown before, all the four images will be saved, for future comparison (Note that if the original image is the same as the original image after rotation, this situation is not repeated ). At the beginning, it was understood that only the original image was saved after several comparisons, WA several times. Method: because the data size is small, the two-dimensional graph is directly saved as a one-dimensional array, and then the violent enumeration can be compared.

The Code is as follows:

# Include <iostream> # include <algorithm> # include <cstring> # include <cstdlib> # include <cstdio> # include <cmath> using namespace std; const int MAXSIZE = 2505; int n, num, cct; char a [52] [52], B [52] [52], c [405] [MAXSIZE]; void Hash () // convert a two-dimensional array to a one-dimensional array and save it {int cct =-1; for (int I = 1; I <= n; I ++) for (int j = 1; j <= n; j ++) c [num] [++ cct] = B [I] [j]; ++ num ;} int Search () {for (int k = 0; k <cct; k ++) {int nn = 0, ff = 1; (Int I = 1; ff & I <= n; I ++) for (int j = 1; ff & j <= n; j ++) if (c [k] [nn ++]! = B [I] [j]) {ff = 0; break;} if (ff) return 1;} return 0;} int main () {# ifdef test freopen ("sample.txt", "r", stdin); # endif int I, N, x [102], y [102]; char str [102]; while (scanf ("% d", & n), n) {int flag = 1; cct = num = 0; memset (a, '0 ', sizeof (a); for (I = 1; I <= n; I ++) a [I] [n + 1] = 0; N = 2 * n; for (I = 1; I <= N; I ++) scanf ("% d % c", & x [I], & y [I], & str [I]); for (I = 1; I <= N; I ++) {if (str [I] = '+ ') a [x [I] [y [I] = '1'; else a [x [I] [y [I] = '0 '; memcpy (B, a, sizeof (a); if (Search () break; Hash (); for (int j = 1; j <= n; j ++) for (int k = 1; k <= n; k ++) B [k] [n-j + 1] = a [j] [k]; if (Search () break; Hash (); for (int j = 1; j <= n; j ++) for (int k = 1; k <= n; k ++) B [n-k + 1] [j] = a [j] [k]; if (Search () break; Hash (); for (int j = 1; j <= n; j ++) for (int k = 1; k <= n; k ++) B [j] [n-k + 1] = a [j] [k]; if (Search () break; Hash (); cct = num ;} if (I % 2) flag = 2; if (I <= N) printf ("Player % d wins on move % d \ n", flag, I ); else puts ("Draw");} return 0 ;}