用C语言实现井字棋

好的，下面是一个简单的井字棋的C语言实现：

#include <stdio.h>

// 定义棋盘大小
#define ROWS 3
#define COLS 3

// 定义棋子类型
#define EMPTY 0
#define X 1
#define O 2

// 初始化棋盘
void init_board(int board[][COLS]) {
    int i, j;
    for (i = 0; i < ROWS; i++) {
        for (j = 0; j < COLS; j++) {
            board[i][j] = EMPTY;
        }
    }
}

// 打印棋盘
void print_board(int board[][COLS]) {
    int i, j;
    for (i = 0; i < ROWS; i++) {
        for (j = 0; j < COLS; j++) {
            if (board[i][j] == X) {
                printf("X ");
            } else if (board[i][j] == O) {
                printf("O ");
            } else {
                printf("- ");
            }
        }
        printf("\n");
    }
}

// 判断是否有玩家获胜
int check_win(int board[][COLS], int player) {
    int i, j;
    for (i = 0; i < ROWS; i++) {
        if (board[i][0] == player && board[i][1] == player && board[i][2] == player) {
            return 1;
        }
    }
    for (j = 0; j < COLS; j++) {
        if (board[0][j] == player && board[1][j] == player && board[2][j] == player) {
            return 1;
        }
    }
    if (board[0][0] == player && board[1][1] == player && board[2][2] == player) {
        return 1;
    }
    if (board[0][2] == player && board[1][1] == player && board[2][0] == player) {
        return 1;
    }
    return 0;
}

// 判断是否平局
int check_tie(int board[][COLS]) {
    int i, j;
    for (i = 0; i < ROWS; i++) {
        for (j = 0; j < COLS; j++) {
            if (board[i][j] == EMPTY) {
                return 0;
            }
        }
    }
    return 1;
}

// 玩家下棋
void player_move(int board[][COLS]) {
    int row, col;
    printf("Enter row and column (1-3): ");
    scanf("%d %d", &row, &col);
    row--;
    col--;
    if (row < 0 || row >= ROWS || col < 0 || col >= COLS || board[row][col] != EMPTY) {
        printf("Invalid move. Try again.\n");
        player_move(board);
    } else {
        board[row][col] = X;
    }
}

// AI下棋
void ai_move(int board[][COLS]) {
    int row, col;
    printf("AI's turn:\n");
    // 简单的AI策略：随机下一个可行的位置
    do {
        row = rand() % ROWS;
        col = rand() % COLS;
    } while (board[row][col] != EMPTY);
    board[row][col] = O;
}

int main() {
    int board[ROWS][COLS];
    init_board(board);
    printf("Welcome to Tic-Tac-Toe!\n");
    printf("You are playing as X, AI is playing as O.\n");
    printf("Enter row and column numbers to make your move.\n");
    printf("Here is the initial board:\n");
    print_board(board);
    while (1) {
        player_move(board);
        print_board(board);
        if (check_win(board, X)) {
            printf("Congratulations! You win!\n");
            break;
        }
        if (check_tie(board)) {
            printf("It's a tie!\n");
            break;
        }
        ai_move(board);
        print_board(board);
        if (check_win(board, O)) {
            printf("AI wins! Better luck next time.\n");
            break;
        }
        if (check_tie(board)) {
            printf("It's a tie!\n");
            break;
        }
    }
    return 0;
}

这个程序实现了一个简单的井字棋游戏，玩家使用 X 表示，AI使用 O 表示。玩家和AI轮流下棋，第一个在一行、一列或一条对角线上连成三个棋子的人获胜，如果棋盘填满了但没有人获胜，则为平局。