c++ - C++井字游戏AI

我的井字游戏几乎完成了。目前，它被设置为两人对人，但我知道我必须实现一个简单的AI才能获得批准。现在，我需要您的帮助。我知道我必须在一些小步骤中考虑一下，例如三种“举动”方法，例如

如果AI在第一列中移动了&&，则打开了右侧的两个框，在其中一个框中进行移动并返回true

如果AI在中间移动了&&，则打开了左右框，请在其中一个框内移动，并返回true

如果AI在第3列中移动了&&，则打开了左侧的两个框，请在其中一个框中进行移动并返回true

我无法完全理解如何在下面的代码中实现它:

#include <iostream>
using namespace std;
char matrix[3][3] = { '7', '8', '9', '4', '5', '6', '1', '2', '3' };
char player = 'X';
int n;
void Draw()
{
    system("cls");
    cout << "Tic Tac Toe !\n" << endl;
    for (int i = 0; i < 3; i++)
    {
        for (int j = 0; j < 3; j++)
        {
            cout << matrix[i][j] << " ";
        }
        cout << endl;
    }
}
void Input()
{
    int a;
    cout << "\nIt's " << player << " turn. " << "Press the number of the field: ";
    cin >> a;

    if (a == 7)
    {
        if (matrix[0][0] == '7')
            matrix[0][0] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 8)
    {
        if (matrix[0][1] == '8')
            matrix[0][1] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 9)
    {
        if (matrix[0][2] == '9')
            matrix[0][2] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 4)
    {
        if (matrix[1][0] == '4')
            matrix[1][0] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 5)
    {
        if (matrix[1][1] == '5')
            matrix[1][1] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 6)
    {
        if (matrix[1][2] == '6')
            matrix[1][2] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 1)
    {
        if (matrix[2][0] == '1')
            matrix[2][0] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 2)
    {
        if (matrix[2][1] == '2')
            matrix[2][1] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }
    else if (a == 3)
    {
        if (matrix[2][2] == '3')
            matrix[2][2] = player;
        else
        {
            cout << "Field is already in use try again!" << endl;
            Input();
        }
    }

}
void TogglePlayer()
{
    if (player == 'X')
        player = 'O';
    else
        player = 'X';
}
char Win()
{
    //first player
    if (matrix[0][0] == 'X' && matrix[0][1] == 'X' && matrix[0][2] == 'X')
        return 'X';
    if (matrix[1][0] == 'X' && matrix[1][1] == 'X' && matrix[1][2] == 'X')
        return 'X';
    if (matrix[2][0] == 'X' && matrix[2][1] == 'X' && matrix[2][2] == 'X')
        return 'X';

    if (matrix[0][0] == 'X' && matrix[1][0] == 'X' && matrix[2][0] == 'X')
        return 'X';
    if (matrix[0][1] == 'X' && matrix[1][1] == 'X' && matrix[2][1] == 'X')
        return 'X';
    if (matrix[0][2] == 'X' && matrix[1][2] == 'X' && matrix[2][2] == 'X')
        return 'X';

    if (matrix[0][0] == 'X' && matrix[1][1] == 'X' && matrix[2][2] == 'X')
        return 'X';
    if (matrix[2][0] == 'X' && matrix[1][1] == 'X' && matrix[0][2] == 'X')
        return 'X';

    //second player
    if (matrix[0][0] == 'O' && matrix[0][1] == 'O' && matrix[0][2] == 'O')
        return 'O';
    if (matrix[1][0] == 'O' && matrix[1][1] == 'O' && matrix[1][2] == 'O')
        return 'O';
    if (matrix[2][0] == 'O' && matrix[2][1] == 'O' && matrix[2][2] == 'O')
        return 'O';

    if (matrix[0][0] == 'O' && matrix[1][0] == 'O' && matrix[2][0] == 'O')
        return 'O';
    if (matrix[0][1] == 'O' && matrix[1][1] == 'O' && matrix[2][1] == 'O')
        return 'O';
    if (matrix[0][2] == 'O' && matrix[1][2] == 'O' && matrix[2][2] == 'O')
        return 'O';

    if (matrix[0][0] == 'O' && matrix[1][1] == 'O' && matrix[2][2] == 'O')
        return 'O';
    if (matrix[2][0] == 'O' && matrix[1][1] == 'O' && matrix[0][2] == 'O')
        return 'O';

    return '/';
}

int main()
{
    n = 0;
    Draw();
    while (1)
    {
        n++;
        Input();
        Draw();
        if (Win() == 'X')
        {
            cout << "X wins!" << endl;
            break;
        }
        else if (Win() == 'O')
        {
            cout << "O wins!" << endl;
            break;
        }
        else if (Win() == '/' && n == 9)
        {
            cout << "It's a draw!" << endl;
            break;
        }
        TogglePlayer();
    }
    system("pause");
    return 0;
}

最佳答案

可以使用Minimax算法来实现诸如Tic-Tac-Toe之类的简单棋盘游戏的计算机玩家，该算法可以使用α-β-pruning进行改进。尽管最终的实现会很小，但是可能需要一些时间来理解。

关于c++ - C++井字游戏AI，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/27925608/