c++ - Connect 4(C++) 中的对角线胜利

Question

我正在用 C++ 制作 connect 4 游戏。我制作的游戏是人类与 AI 的对手，因此对于胜利参数，它会检查是人类获胜还是 AI 获胜。到目前为止，我已经能够为行中的胜利和列中的行执行此操作。我唯一遇到的麻烦是对角连续检测四个。

编辑:我的棋盘是一个 7x6 的网格

这是我检查行的方法:

for (int i = 0; i < board.size(); ++i)
    {
        if ((board[i][0] == PLAYER_TOKEN && board[i][1] == PLAYER_TOKEN && board[i][2] == PLAYER_TOKEN && board[i][3] == PLAYER_TOKEN) ||
            (board[i][1] == PLAYER_TOKEN && board[i][2] == PLAYER_TOKEN && board[i][3] == PLAYER_TOKEN && board[i][4] == PLAYER_TOKEN) ||
            (board[i][2] == PLAYER_TOKEN && board[i][3] == PLAYER_TOKEN && board[i][4] == PLAYER_TOKEN && board[i][5] == PLAYER_TOKEN) ||
            (board[i][3] == PLAYER_TOKEN && board[i][4] == PLAYER_TOKEN && board[i][5] == PLAYER_TOKEN && board[i][6] == PLAYER_TOKEN))
        {
            gameover = true;
            human.setWins(human.getWins()+1);
            cout<<"Congratulations You have Won"<<endl;
            system("PAUSE");
        }
    }

对于专栏我有这个:

for (int j = 0; j < board[0].size(); ++j)
{
    if ((board[0][j] == PLAYER_TOKEN && board[1][j] == PLAYER_TOKEN && board[2][j] == PLAYER_TOKEN && board[3][j] == PLAYER_TOKEN) ||
        (board[1][j] == PLAYER_TOKEN && board[2][j] == PLAYER_TOKEN && board[3][j] == PLAYER_TOKEN && board[4][j] == PLAYER_TOKEN) ||
        (board[2][j] == PLAYER_TOKEN && board[3][j] == PLAYER_TOKEN && board[4][j] == PLAYER_TOKEN && board[5][j] == PLAYER_TOKEN))
    {
        gameover = true;
        human.setWins(human.getWins()+1);
        cout<<"Congratulations You have Won"<<endl;
        system("PAUSE");
    }
}

我一直在尝试找到一种类似的方法来实现对角线胜利，但无法想出任何可行的方法。我所得到的只是这个，坦率地说，它没有任何用处!

//for (int rows = 0; rows<4; ++rows){
//  for (int columns = 0; columns<3; ++columns){
//      if ((board[rows][columns]     == PLAYER_TOKEN && board[rows+1][columns+1]   == PLAYER_TOKEN && board[rows+2][columns+2]     == PLAYER_TOKEN && board[rows+3][columns+3] == PLAYER_TOKEN) || 
//      (board[rows+1][columns+1] == PLAYER_TOKEN && board[rows+2][columns+2] == PLAYER_TOKEN && board[rows+3][columns+3] == PLAYER_TOKEN && board[rows+1][columns+4] == PLAYER_TOKEN) || 
//      (board[rows+2][columns+2] == PLAYER_TOKEN && board[rows+3][columns+3] == PLAYER_TOKEN && board[rows+4][columns+4] == PLAYER_TOKEN && board[rows+5][columns+5] == PLAYER_TOKEN))
//      {
//          gameover = true;
//          human.setWins(human.getWins()+1);
//          cout<<"Congratulations you have Won"<<endl;
//          system("PAUSE");
//      }
//  }
//}

有人可以帮忙吗？

提前致谢。

Answer 1

您必须分别处理每个单元格，并尝试检查是否有从该单元格开始的获胜对角线。这是一个递归的解决方案

bool isWinnerRight(int row, int column, int count) {
  if (row < rows && column < columns && board[row][column] == PLAYERTOKEN) {
    if (count == 4) {
      return true;
    } else {
      return isWinnerRight(row+1.column+1, count+1);
    }
  } else return false;
}

bool isWinnerLeft(int row, int column, int count) {
  if (row < rows && column >= 0 && board[row][column] == PLAYERTOKEN) {
    if (count == 4) {
      return true;
    } else {
      return isWinnerRight(row+1.column-1, count+1);
    }
  } else return false;
}

int main() {
   //do the initialization here
   for (int i = 0; i < rows, i++) {
     for (int j = 0; j < columns; j++) {
       if (isWinnerRight(i,j,0)){/*WINNER FOUND*/}
       if (isWinnerLeft(i,columns-j,0)){/*WINNER FOUND*/}
     }
   }
}

除了在递归调用中，isWinnerLeft() 与 isWinnerRight 类似