C++實現超贊的解魔方的機器人代碼

bcf2 9年前發布 | 1K 次閱讀 C/C++

C++實現超贊的解魔方的機器人代碼,這段代碼精簡實用,作者的腦子不知道是怎么長的,厲害。

/**
 *

  • 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] + isCorner12; int killer = affectedCubies[face][(i&3)==3 ? i-3 : i+1] + isCorner12;; 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)) << (2e); for( int c=0; c<8; c++ ) result[1] |= ((state[c+12]-12) & 5) << (3c); 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: ; } }</pre>

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