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前段时间我用 C++ 制作了一个类似于五子棋的游戏，需要两个玩家参与。

现在我想让它成为 Player vs Computer。我试图以最简单的方式做到这一点，通过使计算机的功能来选择一个随机单元格，但我仍然没有成功。我知道为了得到一个随机数，我可以使用 rand() 和这样的字母：

char letters[] = "abcdefghijklmnopqrstuvwxyz";
char x = letters[rand() % 26];

有人可以帮助我并描述如何实现计算机播放器吗？到目前为止，这是我的实现：

#include <iostream>
#include <iomanip>
using namespace std;

void print_table(int x[][15]) {
    system("cls");
    for (int i = 0; i < 15; i++) {//the loop that use to print out the english character row
        if (i == 0)
            cout << setw(4) << "A";
        else if (i == 1)
            cout << " B";
        else if (i == 2)
            cout << " C";
        else if (i == 3)
            cout << " D";
        else if (i == 4)
            cout << " E";
        else if (i == 5)
            cout << " F";
        else if (i == 6)
            cout << " G";
        else if (i == 7)
            cout << " H";
        else if (i == 8)
            cout << " I";
        else if (i == 9)
            cout << " J";
        else if (i == 10)
            cout << " K";
        else if (i == 11)
            cout << " L";
        else if (i == 12)
            cout << " M";
        else if (i == 13)
            cout << " N";
        else if (i == 14)
            cout << " O";
        else if (i == 15)
            cout << " P";
    }
    cout << endl;
    for (int i = 0; i < 15; i++) {
        cout << setw(2) << i;//print out the row number
        for (int j = 0; j < 15; j++) {//print out the board game.
            if (x[i][j] == 0) {//the inital value is 0, so when the block is 0 then print out the '.'
                cout << " .";
            }
            else if (x[i][j] == 1) {//when the player O input the block then the value will adding one then if check the block is one then output the 'o'
                cout << " O";
            }
            else if (x[i][j] == 2) {//when the player X input the block then the value will adding two then if check the block is two then output the 'x'
                cout << " X";
            }
        }
        cout << endl;
    }
}
int check_player(int p) {
    if (p == 1) {//change the player everytime before the next loop compile
        p++;
    }
    else {
        p--;
    }
    return p;
}
void input_value(int &t, int &n, int p, int x[][15]) {
    char eng;
    int number;
    do {//the loop that ask for the user input the location.
        cout << "player ";
        if (p == 1) {
            cout << "O";
        }
        else {
            cout << "X";
        }
        cout << ", make a move: ";
        cin >> eng;//input the location
        cin >> number;
        if (eng == 'A')//change the character to different number
            t = 0;
        else if (eng == 'B')
            t = 1;
        else if (eng == 'C')
            t = 2;
        else if (eng == 'D')
            t = 3;
        else if (eng == 'E')
            t = 4;
        else if (eng == 'F')
            t = 5;
        else if (eng == 'G')
            t = 6;
        else if (eng == 'H')
            t = 7;
        else if (eng == 'I')
            t = 8;
        else if (eng == 'J')
            t = 9;
        else if (eng == 'K')
            t = 10;
        else if (eng == 'L')
            t = 11;
        else if (eng == 'M')
            t = 12;
        else if (eng == 'N')
            t = 13;
        else if (eng == 'O')
            t = 14;
        if (!(eng >= 'A'&&eng <= 'M') || !(number >= 0 && number <= 14) || x[number][t] != 0) {//when the input wrong, output the statement to ask anouther input and loop again.
            cout << "Invaid input, Try again!" << endl;
            continue;
        }
        else {//if no problem then this input loop is break and jump to the next statement
            break;
        }
    } while (1);//Because it will break as well so the do-while loop is no any requirement
    n = number;
}
int main() {
    const int num = 15;//the number for constant the array row and column value
    char check_e;//for the user input the column
    int R[num][num] = { 0 }, check_n, player = 1, buger = 0, transfer, playerO_win = 0, playerX_win = 0, draw = 0, check_draw;//the variable that for user input or checking the game statment
    do {//launch the loop for the user input again and again
        check_draw = 0;//reset the checking of draw
        print_table(R);
        input_value(transfer, check_n, player, R);
        R[check_n][transfer] += player;//change the value according the player's input and the player name.
        for (int i = 0; i < num; i++) {
            for (int j = 0; j < num; j++) {
                if (i <= 8 && R[j][i] != 0 && (R[j][i] == R[j][i + 1] && R[j][i] == R[j][i + +2] && R[j][i] == R[j][i + 3] && R[j][i] == R[j][i + 4])) {//the checking for the row bingo
                    if (R[j][i] == 1) {
                        playerO_win++;
                        break;
                    }
                    else {
                        playerX_win++;
                        break;
                    }
                }
                else if (j <= 8 && R[j][i] != 0 && (R[j][i] == R[j + 1][i] && R[j][i] == R[j + 2][i] && R[j][i] == R[j + 3][i] && R[j][i] == R[j + 4][i])) {//the checking for the column bingo
                    if (R[j][i] == 1) {
                        playerO_win++;
                        break;
                    }
                    else {
                        playerX_win++;
                        break;
                    }
                }
                else if (j <= 8 && i <= 8 && R[j][i] != 0 && (R[j][i] == R[j + 1][i + 1] && R[j][i] == R[j + 2][i + 2] && R[j][i] == R[j + 3][i + 3] && R[j][i] == R[j + 4][i + 4])) {//the checking for the \ situation.
                    if (R[j][i] == 1) {
                        playerO_win++;
                        break;
                    }
                    else {
                        playerX_win++;
                        break;
                    }
                }
                else if ((j >= 4 || i >= 4 || i <= 8) && R[j][i] != 0 && (R[j][i] == R[j - 1][i + 1] && R[j][i] == R[j - 2][i + 2] && R[j][i] == R[j - 3][i + 3] && R[j][i] == R[j - 4][i + 4])) {//the checking for the / situation
                    if (R[j][i] == 1) {
                        playerO_win++;
                        break;
                    }
                    else {
                        playerX_win++;
                        break;
                    }
                }
                for (int i = 0; i < num; i++) {//the loop for checking the draw
                    for (int j = 0; j < num; j++) {//this loop will check for every time compilation.
                        if (R[j][i] == 0)//when there are any empty block then the check_draw will adding, the draw situation is the check_draw be 0
                            check_draw++;
                    }
                }
                if (check_draw == 0) {//when the check_draw equal to 0 which mean the situation is no empty block
                    draw++;
                    break;
                }

            }
            if (playerO_win != 0 || playerX_win != 0 || draw == 1)//break the second loop
                break;
        }
        if (playerO_win == 1 && playerX_win == 0) {// when the player win print the block game again and print out the win statement
            print_table(R);
            cout << "player O wins!" << endl;
            break;
        }
        else if (playerX_win == 1 && playerO_win == 0) {//the other player win the game
            print_table(R);
            cout << "player X wins!" << endl;
            break;
        }
        else if (draw == 1) {//the draw block game print
            print_table(R);
            cout << "Draw game!" << endl;
            break;
        }
        player = check_player(player);

    } while (1);//in fact it is no need for the loop statement, because most of the situation will have a break statement for out of the loop
    return 0;
}
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1 回答 1

0

以下是我可能会根据您的初始实现来实现它的方式：

#include <iostream>
#include <iomanip>
#include <cstdlib>

using namespace std;

#define SIZE 10
#define LINE_LENGTH 4
#define COMPUTER_O_PLAYER true
#define COMPUTER_X_PLAYER false

void print_board(unsigned char x[SIZE][SIZE]) {
    cout << "   ";
    for (int i = 0; i < SIZE; i++) { // The loop that use to print out the english character row.
        cout << (char) ('A' + i) << " ";
    }
    cout << endl;
    for (int i = 0; i < SIZE; i++) {
        cout << setw(2) << i; // Print out the row number.
        for (int j = 0; j < SIZE; j++) {//print out the board game.
            if (x[i][j] == 0) { // Unoccupied tile.
                cout << " .";
            } else if (x[i][j] == 1) { // The tile belongs to X.
                cout << " X";
            } else { // The tile belongs to O.
                cout << " O";
            }
        }
        cout << endl;
    }
}

void get_position(int &x, int &y, bool player, unsigned char board[SIZE][SIZE]) {
    char eng;

    // The loop that ask for the user input the location.
    do {
        cout << "Player " << (player ? "X" : "O") << ", make a move: ";
        cin >> eng;
        y = toupper(eng) - 'A';
        cin >> x;
        if (!(x >= 0 && x < SIZE) || !(y >= 0 && y < SIZE) || board[x][y] != 0) {
            // When the input wrong, output the statement to ask another input and loop again.
            cout << "Invalid input, Try again!" << endl;
            continue;
        } else { // If no problem then this input loop is break and jump to the next statement.
            break;
        }
    } while (true);
}

void get_random_position(int &x, int &y, bool player, unsigned char board[SIZE][SIZE]) {
    do {
        x = rand() % SIZE;
        y = rand() % SIZE;
    } while ((!(x >= 0 && x < SIZE) || !(y >= 0 && y < SIZE) || board[x][y] != 0));
    cout << "Player " << (player ? "X" : "O") << " chose: " << (char) ('A' + y) << x << endl;
}

unsigned int
count_in_direction(unsigned char board[SIZE][SIZE], unsigned int x, unsigned int y, short int m_x, short int m_y) {
    unsigned int count = 0;
    unsigned char tile = board[x][y];

    while (((x + m_x >= 0) && (x + m_x < SIZE)) && ((y + m_y >= 0) && (y + m_y < SIZE)) &&
           (board[x + m_x][y + m_y] == tile)) {
        x += m_x;
        y += m_y;
        count++;
    }

    return count;
}

bool full_line(unsigned char board[SIZE][SIZE], unsigned int x, unsigned int y) {
    const short int directions[4][2] = {{1, 0},
                                        {1, -1},
                                        {0, 1},
                                        {1, 1}};
    for (const auto &direction : directions) {
        if (LINE_LENGTH - 1 <= (count_in_direction(board, x, y, direction[0], direction[1]) +
                                count_in_direction(board, x, y, -direction[0], -direction[1])))
            return true;
    }

    return false;
}

int main() {
    // The variable that for user input or checking the game state.
    unsigned char board[SIZE][SIZE] = {0};
    int x, y;
    bool player = true;
    bool draw;

    srand(time(nullptr));

    // Run the game.
    do {
        print_board(board);
        if ((player && COMPUTER_X_PLAYER) || (!player && COMPUTER_O_PLAYER)) {
            get_random_position(x, y, player, board);
        } else {
            get_position(x, y, player, board);
        }
        board[x][y] = player ? 1 : 2;
        if (full_line(board, x, y)) {
            print_board(board);
            cout << "player " << (player ? "X" : "O") << " wins!" << endl;
            break;
        }

        draw = true;
        for (int k = 0; (k < SIZE) && draw; k++) {
            for (int l = 0; (l < SIZE) && draw; l++) {
                if (board[k][l] == 0) draw = false;
            }
        }

        if (draw) {
            print_board(board);
            cout << "Draw game!" << endl;
            break;
        }
        player = !player;

    } while (true);

    return 0;
}

请深入了解此实现并内化其方法，它还支持作为用户、用户和计算机的玩家，以及同时作为计算机的玩家。

一些见解：

  • if ... else当有更简单的方法时，不要使用 long 。
  • 对变量使用正确的数据类型。
  • 使用指示性变量命名。
  • 尽可能使用循环，不要 long if ... else。
  • 保持代码格式正确。
  • 用于#define定义要使用的全局常量，以便以后更容易更改和可读性。
  • 您确实可以使用rand，但您不需要获取字符，您可以立即获取索引并避免转换。
于 2020-11-24T19:37:56.960 回答