上的N×N大小板

井字得獎有由N個井字板的N，用2名選手。本場比賽是正常井字不同，因爲比賽只結束時，董事會完全填補，一個點是在一排一排一排授予三，二分數是四，三五年，和等等。「連續」可以表示水平，垂直或對角線。給定一個完成董事會，確定由誰儘可能高效具有最高分的獲得者。上的N×N大小板

這是我完成的代碼。我想知道是否有更好的方法來做到這一點。只是改變了「N」的數量在全局變量在頂部，你應該能夠與任何電路板尺寸運行。

const int boardSize = N; 

//used for calculating all points except for diagonals with a negative slope 
void counter(vector<<vector>> &ticTacToe, int &score1, int &score2, int wid, bool horz, bool diag) { 
    int count = 0, hei = 0; 
    if (horz != diag)////allows us to traverse the board vertically instead of horizontally 
     swap(wid, hei); 
    while (hei < boardSize && wid < boardSize) { 
     while (hei < boardSize && wid < boardSize && ticTacToe[wid][hei]) { 
      count++; 
      if (horz) { 
       wid++; 
       if (diag)//if traversing diagonally, need to increment both width & height 
        hei++; 
      } 
      else { 
       hei++; 
       if (diag)//if traversing diagonally, need to increment both width & height 
        wid++; 
      } 
     } 
     if (count > 2) { 
      score1 += (count - 2); 
     } 
     count = 0; 

     //only difference from above is ticTacToe[wid][hei] 
     while (hei < boardSize && wid < boardSize && !ticTacToe[wid][hei]) { 
      count++; 
      if (horz) { 
       wid++; 
       if (diag) 
        hei++; 
      } 
      else { 
       hei++; 
       if (diag) 
        wid++; 
      } 
     } 
     if (count > 2) { 
      score2 += (count - 2); 
     } 
     count = 0; 
    } 
    if (horz !& diag) { //should change? 
     swap(wid, hei); 
    } 
} 


//used for calculating diagonal points with a diagonal slope 
void negSlopeDiagonal(vector<vector<int>> &ticTacToe, int &score1, int &score2, int wid, bool horz) { 
    int count = 0, hei = boardSize - 1; 
    if (!horz) { 
     hei -= wid; 
     wid = 0; 
    } 
    while (wid < boardSize && hei >= 0) { 
     while (wid < boardSize && hei >= 0 && ticTacToe[wid][hei]) { 
      count++; 
      hei--; 
      wid++; 
     } 
     if (count > 2) { 
      score1 += (count - 2); 
     } 
     count = 0; 
     while (wid < boardSize && hei >= 0 && !ticTacToe[wid][hei]) { 
      count++; 
      hei--; 
      wid++; 
     } 
     if (count > 2) { 
      score_P2 += (count - 2); 
     } 
     count = 0; 
    } 
} 


bool winner(vector<vector<int>> ticTacToe) { 
    int score_P1 = 0, score_P2 = 0; 

    //count points vertically 
    for (int wid = 0; wid < boardSize; wid++) { 
     counter(ticTacToe, score_P1, score_P2, wid, false, false); 
    } 

    //count points horizontally 
    for (int wid = 0; wid < boardSize; wid++) { 
     counter(ticTacToe, score_P1, score_P2, wid, true, false); 
    } 

    //counts diagonal points sloping up going vertical 
    for (int wid = 0; wid < boardSize; wid++) { 
     counter(ticTacToe, score_P1, score_P2, wid, false, true); 
    } 

    //counts diagonal points sloping up going horizontal 
    for (int wid = 1; wid < boardSize; wid++) { 
     counter(ticTacToe, score_P1, score_P2, wid, true, true); 
    } 

    //counts diagonal points sloping down going horizonatal 
    for (int wid = 0; wid < boardSize; wid++) { 
     negSlopeDiagonal(ticTacToe, score_P1, score_P2, wid, true); 
    } 

    //counts diagonal points sloping down going vertical 
    for (int wid = 1; wid < boardSize; wid++) { 
     negSlopeDiagonal(ticTacToe, score_P1, score_P2, wid, false); 
    } 
    cout << "P1: " << score_P1 << endl << "P2: " << score_P2 << endl; 
    return (score_P1 > score_P2); 
} 


void printBoard(const vector<vector<int>> gameboard) { 
    assert(gameboard.size() == boardSize && gameboard[0].size() == boardSize); 
    for (int j = boardSize - 1; j >= 0; j--) { 
     for (int i = 0; i < boardSize; i++) { 
      cout << gameboard[i][j] << " "; 
     } 
     cout << endl; 
    } 
} 

int main() { 
    vector<int> empt; //empty vector for initialization 
    vector<vector<int>> board(boardSize, empt); 

    for (int i = 0; i < boardSize; i++) { 
     for (int j = 0; j < boardSize; j++) { 
      board[i].push_back((bool)(rand() % 2)); 
     } 
    } 
    printBoard(board); 
    if(winner(board)) { 
     cout << "P1 wins!\n"; 
    } 
    else { 
     cout << "P2 wins!\n"; 
    } 
    return 0; 
}

來源

2016-03-09 Mitch D

您可以使用方向矢量來遍歷棋盤。例如：

int traverse_and_calculate(const vector< vector<int> > &board, int start_row, int start_column, int inc_row, int inc_col, int length) 
{ 
    for (int row = start_row, column = start_column, step = 0; step < length; row += inc_row, col += inc_col) 
    { 

    } 
}

並以這種方式，你可以輕鬆地處理所有情況中的一個或兩個環。

來源

2016-03-09 05:51:20

上的N×N大小板

回答

相關問題