/* ** *** @üťhöř : Řëwä Mąťhúŕ @Śťaťüš : Sťïľľ Ïñ Çöñšťŕūçťīöñ @Ďäťë : 18 Jäñ 2018 ** *** */ /* // this codes uses some concepts of vector and dynamic allocation to make an pattern //***uses the dynamic and backtracking approach for making an pattern which is looks like an piston. //iterators are used with the pattern generation are used in the creation of code so that it can leads to learn some concepts any query about the code make comment by pinging me will try to solve your query // Pardon me i Will soon make some comments to illustrate all this code processing. *///**** //***** #include #include #include #include #include #include using namespace std; // defined an structure of nodes for dynamic graphics. struct node { int val; unsigned char neighbors; }; //patten class is defined for using the variable methods and class members for generating the pattern. class nPattern { //public membor function are defined to getting the path values for the pattern public: nPattern() { dx[0] = -1; dy[0] = -2; dx[1] = -1; dy[1] = 2; dx[2] = 1; dy[2] = -2; dx[3] = 1; dy[3] = 2; dx[4] = -2; dy[4] = -1; dx[5] = -2; dy[5] = 1; dx[6] = 2; dy[6] = -1; dx[7] = 2; dy[7] = 1; } // to made an solution of pattern i have created the solution function to find the pattern with vector variables in which width and height of the pattern is set and used for generation of pattern. void solve( vector& puzz, int max_wid ) { if( puzz.size() < 1 ) return; wid = max_wid; hei = static_cast( puzz.size() ) / wid; int len = wid * hei, c = 0; max = len; arr = new node[len]; memset( arr, 0, len * sizeof( node ) ); // to iterate the pattern for the graph path to create the pattern in an unique way. for( vector::iterator i = puzz.begin(); i != puzz.end(); i++ ) { if( ( *i ) == "*" ) { max--; arr[c++].val = -1; continue; } arr[c].val = atoi( ( *i ).c_str() ); c++; } // to free the memory after using the memory space after the pattern made. solveIt(); c = 0; for( vector::iterator i = puzz.begin(); i != puzz.end(); i++ ) { if( ( *i ) == "." ) { ostringstream o; o << arr[c].val; ( *i ) = o.str(); } c++; } delete [] arr; } //for searching an way to find the numbers for the pattern balancing. private: bool search( int x, int y, int w ) { if( w > max ) return true; //find the neighbour of that number by using dynamic and backtracking approach. node* n = &arr[x + y * wid]; n->neighbors = getNeighbors( x, y ); for( int d = 0; d < 8; d++ ) { if( n->neighbors & ( 1 << d ) ) { int a = x + dx[d], b = y + dy[d]; if( arr[a + b * wid].val == 0 ) { arr[a + b * wid].val = w; if( search( a, b, w + 1 ) ) return true; arr[a + b * wid].val = 0; } } } return false; } //for printing the neighbour and print it. unsigned char getNeighbors( int x, int y ) { unsigned char c = 0; int a, b; for( int xx = 0; xx < 8; xx++ ) { a = x + dx[xx], b = y + dy[xx]; if( a < 0 || b < 0 || a >= wid || b >= hei ) continue; if( arr[a + b * wid].val > -1 ) c |= ( 1 << xx ); } return c; } void solveIt() { int x, y, z; findStart( x, y, z ); if( z == 99999 ) { cout << "\nCan't find start point!\n"; return; } search( x, y, z + 1 ); } //finding the start point of the path of the pattern. void findStart( int& x, int& y, int& z ) { z = 99999; for( int b = 0; b < hei; b++ ) for( int a = 0; a < wid; a++ ) if( arr[a + wid * b].val > 0 && arr[a + wid * b].val < z ) { x = a; y = b; z = arr[a + wid * b].val; } } int wid, hei, max, dx[8], dy[8]; node* arr; }; int main( int argc, char* argv[] ) { int wid; string p; //p = "* . . . * * * * * . * . . * * * * . . . . . . . . . . * * . * . . * . * * . . . 1 . . . . . . * * * . . * . * * * * * . . . * *"; wid = 8; p = "* * * * * 1 * . * * * * * * * * * * . * . * * * * * * * * * . . . . . * * * * * * * * * . . . * * * * * * * . * * . * . * * . * * . . . . . * * * . . . . . * * . . * * * * * . . * * . . . . . * * * . . . . . * * . * * . * . * * . * * * * * * * . . . * * * * * * * * * . . . . . * * * * * * * * * . * . * * * * * * * * * * . * . * * * * * "; wid = 13; istringstream iss( p ); vector puzz; copy( istream_iterator( iss ), istream_iterator(), back_inserter >( puzz ) ); nPattern s; s.solve( puzz, wid ); int c = 0; for( vector::iterator i = puzz.begin(); i != puzz.end(); i++ ) { if( ( *i ) != "*" && ( *i ) != "." ) { //atoi() use is not considered as safe so I will soon replace it with more stable function strtoi(). if( atoi( ( *i ).c_str() ) < 10 ) cout << "0"; cout << ( *i ) << " "; } else cout << " "; if( ++c >= wid ) { cout << endl; c = 0; } } cout << endl << endl; return system( "pause" ); }

Boring Number Pattern looks like an piston