C++实现超赞的解魔方的机器人代码
2018-07-20 来源:open-open
C++实现超赞的解魔方的机器人代码,这段代码精简实用,作者的脑子不知道是怎么长的,厉害。
/********************************************************************** * * A cube 'state' is a vector<int> with 40 entries, the first 20 * are a permutation of {0,...,19} and describe which cubie is at * a certain position (regarding the input ordering). The first * twelve are for edges, the last eight for corners. * * The last 20 entries are for the orientations, each describing * how often the cubie at a certain position has been turned * counterclockwise away from the correct orientation. Again the * first twelve are edges, the last eight are corners. The values * are 0 or 1 for edges and 0, 1 or 2 for corners. * * http://www.sharejs.com **********************************************************************/ #include <iostream> #include <string> #include <vector> #include <map> #include <queue> #include <algorithm> using namespace std; //---------------------------------------------------------------------- typedef vector<int> vi; //---------------------------------------------------------------------- int applicableMoves[] = { 0, 262143, 259263, 74943, 74898 }; // TODO: Encode as strings, e.g. for U use "ABCDABCD" int affectedCubies[][8] = { { 0, 1, 2, 3, 0, 1, 2, 3 }, // U { 4, 7, 6, 5, 4, 5, 6, 7 }, // D { 0, 9, 4, 8, 0, 3, 5, 4 }, // F { 2, 10, 6, 11, 2, 1, 7, 6 }, // B { 3, 11, 7, 9, 3, 2, 6, 5 }, // L { 1, 8, 5, 10, 1, 0, 4, 7 }, // R }; vi applyMove ( int move, vi state ) { int turns = move % 3 + 1; int face = move / 3; while( turns-- ){ vi oldState = state; for( int i=0; i<8; i++ ){ int isCorner = i > 3; int target = affectedCubies[face][i] + isCorner*12; int killer = affectedCubies[face][(i&3)==3 ? i-3 : i+1] + isCorner*12;; int orientationDelta = (i<4) ? (face>1 && face<4) : (face<2) ? 0 : 2 - (i&1); state[target] = oldState[killer]; //state[target+20] = (oldState[killer+20] + orientationDelta) % (2 + isCorner); state[target+20] = oldState[killer+20] + orientationDelta; if( !turns ) state[target+20] %= 2 + isCorner; } } return state; } int inverse ( int move ) { return move + 2 - 2 * (move % 3); } //---------------------------------------------------------------------- int phase; //---------------------------------------------------------------------- vi id ( vi state ) { //--- Phase 1: Edge orientations. if( phase < 2 ) return vi( state.begin() + 20, state.begin() + 32 ); //-- Phase 2: Corner orientations, E slice edges. if( phase < 3 ){ vi result( state.begin() + 31, state.begin() + 40 ); for( int e=0; e<12; e++ ) result[0] |= (state[e] / 8) << e; return result; } //--- Phase 3: Edge slices M and S, corner tetrads, overall parity. if( phase < 4 ){ vi result( 3 ); for( int e=0; e<12; e++ ) result[0] |= ((state[e] > 7) ? 2 : (state[e] & 1)) << (2*e); for( int c=0; c<8; c++ ) result[1] |= ((state[c+12]-12) & 5) << (3*c); for( int i=12; i<20; i++ ) for( int j=i+1; j<20; j++ ) result[2] ^= state[i] > state[j]; return result; } //--- Phase 4: The rest. return state; } //---------------------------------------------------------------------- int main ( int argc, char** argv ) { //--- Define the goal. string goal[] = { "UF", "UR", "UB", "UL", "DF", "DR", "DB", "DL", "FR", "FL", "BR", "BL", "UFR", "URB", "UBL", "ULF", "DRF", "DFL", "DLB", "DBR" }; //--- Prepare current (start) and goal state. vi currentState( 40 ), goalState( 40 ); for( int i=0; i<20; i++ ){ //--- Goal state. goalState[i] = i; //--- Current (start) state. string cubie = argv[i+1]; while( (currentState[i] = find( goal, goal+20, cubie ) - goal) == 20){ cubie = cubie.substr( 1 ) + cubie[0]; currentState[i+20]++; } } //--- Dance the funky Thistlethwaite... while( ++phase < 5 ){ //--- Compute ids for current and goal state, skip phase if equal. vi currentId = id( currentState ), goalId = id( goalState ); if( currentId == goalId ) continue; //--- Initialize the BFS queue. queue<vi> q; q.push( currentState ); q.push( goalState ); //--- Initialize the BFS tables. map<vi,vi> predecessor; map<vi,int> direction, lastMove; direction[ currentId ] = 1; direction[ goalId ] = 2; //--- Dance the funky bidirectional BFS... while( 1 ){ //--- Get state from queue, compute its ID and get its direction. vi oldState = q.front(); q.pop(); vi oldId = id( oldState ); int& oldDir = direction[oldId]; //--- Apply all applicable moves to it and handle the new state. for( int move=0; move<18; move++ ){ if( applicableMoves[phase] & (1 << move) ){ //--- Apply the move. vi newState = applyMove( move, oldState ); vi newId = id( newState ); int& newDir = direction[newId]; //--- Have we seen this state (id) from the other direction already? //--- I.e. have we found a connection? if( newDir && newDir != oldDir ){ //--- Make oldId represent the forwards and newId the backwards search state. if( oldDir > 1 ){ swap( newId, oldId ); move = inverse( move ); } //--- Reconstruct the connecting algorithm. vi algorithm( 1, move ); while( oldId != currentId ){ algorithm.insert( algorithm.begin(), lastMove[ oldId ] ); oldId = predecessor[ oldId ]; } while( newId != goalId ){ algorithm.push_back( inverse( lastMove[ newId ] )); newId = predecessor[ newId ]; } //--- Print and apply the algorithm. for( int i=0; i<(int)algorithm.size(); i++ ){ cout << "UDFBLR"[algorithm[i]/3] << algorithm[i]%3+1; currentState = applyMove( algorithm[i], currentState ); } //--- Jump to the next phase. goto nextPhasePlease; } //--- If we've never seen this state (id) before, visit it. if( ! newDir ){ q.push( newState ); newDir = oldDir; lastMove[ newId ] = move; predecessor[ newId ] = oldId; } } } } nextPhasePlease: ; } }
版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点!
本站所提供的图片等素材,版权归原作者所有,如需使用,请与原作者联系。
上一篇: Java实现基数排序
下一篇:android文件下载代码
最新资讯
热门推荐