我差不多完成了我的tic tac toe游戏。目前它被设置为双人对人,但我知道我必须实施一个简单的AI才能获得批准。现在我需要你的帮助。我知道我必须以三个小步骤来思考它,然后采取行动。像
这样的方法我无法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;
}
答案 0 :(得分:2)
可以使用Minimax算法实现像Tic-Tac-Toe这样的简单棋盘游戏的计算机播放器,可以使用α-β-pruning进行改进。虽然最终的实现将非常小,但可能需要一些时间才能理解。
答案 1 :(得分:-1)
#include <iostream>
#include <string>
using namespace std;
int main()
{
/*
the chart is :
X 2 X
4 5 6
X 8 X
*/
char Matrix[3][3] = { 'X','2','X','4','5','6','X','8','X' };
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
cout << Matrix[i][j] << " ";
}
cout << endl;
}
/*
after this for :
X O X
4 5 6
X O X
*/
int l = 0, m = 0, p[3] = { 0,0,0 };
for (l; l <= 2; l++)
{
for (m; m <= 2; m++)
{
if ((Matrix[l][m]) == 'X')
{
p[l]++;
if ((p[l]) == 2)
{
for (m; m >= 0; m--)
{
if ((Matrix[l][m]) != 'X')
{
Matrix[l][m] = 'O';
}
}
}
}
}
}
return 0;
}
why this code dosent work?