算法设计与分析题目练习四：井字棋（启发式算法）

井字棋又叫做三连棋或一连棋。顾名思义就是让三颗棋子连成一条线就获胜了。国外也有相似的玩法，名字叫tic-tac-toe

#include  
#include  
using namespace std;

/************************************************************************/
/*
	井字棋又叫做三连棋或一连棋。顾名思义就是让三颗棋子连成一条线就获胜了。
	国外也有相似的玩法，名字叫tic-tac-toe
*/
/************************************************************************/
const int MAX_MOVES = 9;
void printBoard(char board[3][3]);
void getNextMove(char board[3][3]);
void win(char board[3][3]);
void whowin(char key);
int foot = 0;

int main()
{
	int i;
	int n = 0;
	char board[3][3] = { '7', '8', '9', '4', '5', '6', '1', '2', '3' };

	printBoard(board);

	for (i = 0; i < MAX_MOVES; i++)
	{
		getNextMove(board);
		win(board);
		n++;
	}

	if (n = MAX_MOVES)
		cout << "gameover";

	system("PAUSE");
	return 0;
}
void win(char a[3][3])
{
	int i, j, k;
	char key;
	if ((a[0][0] == a[1][1] && a[1][1] == a[2][2] && a[2][2] == a[0][0]) || (a[2][0] == a[1][1] && a[1][1] == a[0][2] && a[2][0] == a[0][2]))
	{
		key = a[0][0];
		whowin(key);
	}
	else
	{
		for (i = 0; i < 3; i++)
			for (j = 0; j < 3; j++)
				if (a[i][j] == 15 || a[i][j] == 'X')
				{
					key = a[i][j];
					for (k = 0; k < 3; k++)
					{
						if (a[i][k] != key)
							break;
					}
					if (k == 3)
						whowin(key);
					for (k = 0; k < 3; k++)
					{
						if (a[k][j] != key)
							break;
					}
					if (k == 3)
						whowin(key);
				}
	}
}
void whowin(char key)
{
	if (key == 15)
		cout << "P1 Win!" << endl;
	else
		cout << "P2 Win!" << endl;
	exit(0);
}
void printBoard(char board[3][3])
{
	cout << board[0][0] << " | " << board[0][1] << " | " << board[0][2] << endl;
	cout << "--+---+--" << endl;
	cout << board[1][0] << " | " << board[1][1] << " | " << board[1][2] << endl;
	cout << "--+---+--" << endl;
	cout << board[2][0] << " | " << board[2][1] << " | " << board[2][2] << endl;
}

void getNextMove(char board[3][3])
{
	int i, j, s, t;
	char m;
	while (1)
	{
		s = -1;
		t = -1;
		cout << "Make your move: ";
		cin >> m;
		for (i = 0; i < 3; i++)
			for (j = 0; j < 3; j++)
				if (board[i][j] == m)
				{
					s = i;
					t = j;
				}
		if (s == -1 && t == -1)
			cout << "that point already has a pawn!" << endl;
		else
			break;
	}
	if (foot % 2 == 0)
		board[s][t] = 15;
	else
		board[s][t] = 'X';
	foot++;
	system("cls");
	printBoard(board);
}