Question

有一个N by N tic tac toe board，有2名玩家。这个游戏与普通的tic tac toe不同，因为游戏只在棋盘完全填满时结束。连续三个获得一分，连续四个获得两分，连续五个获得三分，等等。＆＃34;连续＆＃34;可以指水平，垂直或对角线。给定一个完成的董事会，尽可能有效地确定获胜者得分最多的赢家。

这是我完成的代码。我想知道是否有更好的方法来做到这一点。只需更改“N＆＃39; N＆＃39;到顶部的全局变量中的数字，您应该可以使用任何板尺寸运行它。

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;
}

Answer 1

您可以使用方向矢量来遍历电路板。例如：

text.setKeyListener((KeyListener) textView.getTag());

通过这种方式，您可以轻松处理一个或两个循环中的所有情况。

NxN尺寸板上的Tic Tac Toe获胜者

1 个答案: