Usaco magic squares solution report

In the detection 8! = 40320, I found that space is not an influence factor. As long as no repeated operations are performed, the time and space are sufficient.

For reference:

1. Processing of conversions:

/* The set of transformations, in order */

Static int tforms [3] [8] = {8, 7, 6, 5, 4, 3, 2, 1}, {4, 1, 2, 3, 6, 7, 8, 5 },

{1, 7, 2, 4, 5, 3, 6, 8 }};

Here is another way of thinking, which is easier to handle. The idea is to directly understand the conversion as a ing of Different Copy locations (positions after initial position conversion. For example, in the first conversion, 0th is to copy 7th bits (8-1 ). The first conversion is actually this array, that is, the ing of the copy position.

2. Expand conto

Conto expansion is a good theorem: it is used to find a sequence that is arranged, for example, 45213 in the full arrangement of the five numbers. Can directly move here: http://blog.csdn.net/fairyroad/article/details/7555773

CodeWell written.

/* ID: thestor1lang: C ++ task: msquare */# include <iostream> # include <cmath> # include <cstdio> # include <cstring> # include <vector> # include <cassert> # include <string> # include <algorithm> # include <stack> # include <set> using namespace STD; const int size = 8; const int max = 40320; // 8! = 40320 struct square {int h; int square [size]; char op; int Prev; Square () {H =-1;} int Hash () {If (h> 0) {return h;} int h = 0; For (INT I = 0; I <size; ++ I) {H = (H <3) + square [I];} return H ;}}; square squares [Max]; bool equal (square & S1, square & S2) {return s1.hash () = s2.hash ();} bool equal (int * S1, int * S2) {for (INT I = 0; I <size; ++ I) {If (S1 [I]! = S2 [I]) {return false;} return true;} void copy (int * s, int * D) {for (INT I = 0; I <size; ++ I) {d [I] = s [I] ;}} void a (int * Square) {for (INT I = 0; I <= 3; ++ I) {int TMP = square [I]; Square [I] = square [7-I]; Square [7-I] = TMP ;}} void B (int * Square) {int TMP = square [3]; for (INT I = 3; I> = 1; -- I) {square [I] = square [I-1];} square [0] = TMP; TMP = square [4]; for (INT I = 4; I <= 6; ++ I) {square [I] = square [I + 1];} square [7] = TMP;} void C (int * Square) {int TMP = square [1]; Square [1] = square [6]; square [6] = square [5]; Square [5] = square [2]; Square [2] = TMP;} int main () {file * fin = fopen ("msquare. in "," R "); file * fout = fopen (" msquare. out "," W "); // freopen (" log.txt "," W ", stdout); Square targetsq; // int target [size]; for (INT I = 0; I <size; ++ I) {fscanf (FIN, "% d", & targetsq. square [I]);} set <int> SQS; int Top = 0; (INT I = 0; I <size; ++ I) {squares [Top]. square [I] = I + 1;} SQS. insert (squares [Top]. hash (); top ++; int final =-1; for (INT I = 0; I <top; ++ I) {// fprintf (stdout, "I: % d, top: % d \ n ", I, top); Square sq = squares [I]; If (equal (SQ, targetsq) {final = I; break;} copy (sq. square, squares [Top]. square); A (squares [Top]. square); If (SQS. find (squares [Top]. hash () = SQS. end () {squares [Top]. prev = I; squares [Top]. OP = 'a '; SQS. insert (squares [Top]. hash (); top ++;} copy (sq. square, squares [Top]. square); B (squares [Top]. square); If (SQS. find (squares [Top]. hash () = SQS. end () {squares [Top]. prev = I; squares [Top]. OP = 'B'; SQS. insert (squares [Top]. hash (); top ++;} copy (sq. square, squares [Top]. square); C (squares [Top]. square); If (SQS. find (squares [Top]. hash () = SQS. end () {squares [Top]. prev = I; squares [Top]. OP = 'C'; SQS. insert (squares [Top]. Hash (); top ++ }}stack <char> seq; while (final! = 0) {seq. push (squares [Final]. OP); Final = squares [Final]. prev;} fprintf (fout, "% d \ n", seq. size (); While (! Seq. empty () {fprintf (fout, "% C", seq. top (); seq. pop ();} fprintf (fout, "\ n"); Return 0 ;}