这是 Lv1 电脑级别,其中的逻辑是剩余的空位随机选择下在哪里。
程序
//This program is a simple tic-tac-toe game.
#include <iostream>
#include <string>
#include <cstddef>
#include <stdexcept>
#include <ctime>
using namespace std;
char point_now[3][3] = {
' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ' };
char player = 'X';
char computer_player = 'X';
unsigned step_count = 0;
int win_lose(char point[3][3], int n)
{
if (point[0][0] == point[0][1] && point[0][1] == point[0][2] && point[0][0] == 'X') return 1; // X wins
if (point[1][0] == point[1][1] && point[1][1] == point[1][2] && point[1][0] == 'X') return 1; // X wins
if (point[2][0] == point[2][1] && point[2][1] == point[2][2] && point[2][0] == 'X') return 1; // X wins
if (point[0][0] == point[1][0] && point[1][0] == point[2][0] && point[0][0] == 'X') return 1; // X wins
if (point[0][1] == point[1][1] && point[1][1] == point[2][1] && point[0][1] == 'X') return 1; // X wins
if (point[0][2] == point[1][2] && point[1][2] == point[2][2] && point[0][2] == 'X') return 1; // X wins
if (point[0][0] == point[1][1] && point[1][1] == point[2][2] && point[1][1] == 'X') return 1; // X wins
if (point[0][2] == point[1][1] && point[1][1] == point[2][0] && point[1][1] == 'X') return 1; // X wins
if (point[0][0] == point[0][1] && point[0][1] == point[0][2] && point[0][0] == '0') return 2; // 0 wins
if (point[1][0] == point[1][1] && point[1][1] == point[1][2] && point[1][0] == '0') return 2; // 0 wins
if (point[2][0] == point[2][1] && point[2][1] == point[2][2] && point[2][0] == '0') return 2; // 0 wins
if (point[0][0] == point[1][0] && point[1][0] == point[2][0] && point[0][0] == '0') return 2; // 0 wins
if (point[0][1] == point[1][1] && point[1][1] == point[2][1] && point[0][1] == '0') return 2; // 0 wins
if (point[0][2] == point[1][2] && point[1][2] == point[2][2] && point[0][2] == '0') return 2; // 0 wins
if (point[0][0] == point[1][1] && point[1][1] == point[2][2] && point[1][1] == '0') return 2; // 0 wins
if (point[0][2] == point[1][1] && point[1][1] == point[2][0] && point[1][1] == '0') return 2; // 0 wins
if (n == 9) return 3; // end up in a draw
else return 0; // unfinished
}
void game_player_change(char& player)
{
if (player == 'X')
player = '0'; // X -> 0
else player = 'X';// 0 -> X
}
string computer1(char p[3][3])
{
string ret;
const string location_computer[3][3] = {
{
"A1","A2","A3"},{
"B1","B2","B3"},{
"C1","C2","C3"} };
unsigned available_n = 49;
for (int c_i = 0; c_i != 3; c_i++)
{
for (int c_j = 0; c_j != 3; c_j++)
{
if ((p[c_i][c_j] != 'X') && (p[c_i][c_j] != '0'))
p[c_i][c_j] = available_n++; // mark empty places with numbers 1,2,3...
}
}
srand((unsigned)time(NULL));
int ran = rand() % (available_n - 49) + 49; // generate a random number
for (int c_i = 0; c_i != 3; c_i++)
{
for (int c_j = 0; c_j != 3; c_j++)
{
if (p[c_i][c_j] == ran)
ret = location_computer[c_i][c_j]; // the chosen place
}
}
for (int c_i = 0; c_i != 3; c_i++)
{
for (int c_j = 0; c_j != 3; c_j++)
{
if ((p[c_i][c_j] != 'X') && (p[c_i][c_j] != '0'))
p[c_i][c_j] = ' '; // return to Space
}
}
return ret; // this is the copmuter-chosen location
}
int main()
{
std::cout << "This program is a simple tic-tac-toe game.\nProgrammer:Teddy van Jerry\n" << endl;
std::cout << "Difficulty :Level 1.0\n" << endl;
std::cout << "You go first or the computer? You(1), Computer(2): ";
char You_or_Computer;
cin >> You_or_Computer;
cout << endl;
if (You_or_Computer == '1')
computer_player = '0';
else computer_player = 'X';
// Print an empty board.
std::cout << "| |1|2|3|" << endl;
std::cout << "|A| | | |" << endl;
std::cout << "|B| | | |" << endl;
std::cout << "|C| | | |\n" << endl;
while (win_lose(point_now, step_count) == 0)
{
string location;
unsigned location_letter = 0; // A/B/C
unsigned location_number = 0; // 1/2/3
begin: // a label
std::cout << "Player " << player << ", make your move: ";
if (player != computer_player)
cin >> location; // input the location (e.g. B2)
else
{
location = computer1(point_now);
cout << location << endl;
}
std::cout << endl;
switch (location[0])
{
case 'a': case 'A':
location_letter = 0;
break;
case 'b': case 'B':
location_letter = 1;
break;
case 'c': case 'C':
location_letter = 2;
break;
default: // illegal input
location_letter = 100; // indicate an error
break;
}
switch (location[1])
{
case '1':
location_number = 0;
break;
case '2':
location_number = 1;
break;
case '3':
location_number = 2;
break;
default: // illegal input
location_number = 100; // indicate an error
break;
}
try
{
if (location_letter != 100 && location_number != 100 && point_now[location_letter][location_number] == ' ')
point_now[location_letter][location_number] = player;
else throw runtime_error("Illegal input!");
}
catch (runtime_error err)
{
std::cout << err.what() << "\nTry Again? Enter Y or N." << endl;
char decision;
cin >> decision;
if (!cin || decision == 'n' || decision == 'N')
break;
else
{
cout << endl;
goto begin; // go back to the label 'begin'
}
}
// Print the board
std::cout << "| |1|2|3|" << endl;
std::cout << "|A|" << point_now[0][0] << "|" << point_now[0][1] << "|" << point_now[0][2] << "|" << endl;
std::cout << "|B|" << point_now[1][0] << "|" << point_now[1][1] << "|" << point_now[1][2] << "|" << endl;
std::cout << "|C|" << point_now[2][0] << "|" << point_now[2][1] << "|" << point_now[2][2] << "|" << endl;
game_player_change(player); // change the player
++step_count; // count one more time
std::cout << endl;
}
int final_result = win_lose(point_now, step_count);
char winner='N';
switch (final_result)
{
case 1:
winner = 'X';
std::cout << "Congratulations! The winner is X." << endl;
break;
case 2:
winner = '0';
std::cout << "Congratulations! The winner is 0." << endl;
break;
case 3:
std::cout << "The game ended in a draw." << endl;
break;
}
if (winner == computer_player)
cout << "Computer won!" << endl;
else
{
if (winner != 'N')
cout << "You won!!!" << endl;
}
std::cout << "\nALL RIGHTS RESERVED (c) 2020 Teddy van Jerry" << endl;
return 0;
}
//Copyright :2020 Teddy van Jerry
输出示例
真够笨的,哈哈!!!这么轻松就赢了。
分析
- 本来用的
uniform_int_distribution<unsigned> u<m,n>;
时而可以,时而报错。而且随机数每次一样,于是就做出了修改。 - 该程序一样对于不合法输入进行了判别(分为两种:1.坐标超出范围,2.输入到原来已有棋子的位置上),详见【C++ 程序】 井字棋游戏(人 VS 人)中的分析。
- 格外应注意69-76行的内容非常必要,对于全局变量的影响不会随着该函数结束而结束。详见我的博客 关于 C++中 定义函数造成影响 的思考。
ALL RIGHTS RESERVED © 2020 Teddy van Jerry
欢迎转载,转载请注明出处。
See also
Teddy van Jerry 的导航页
【C++ 程序】 井字棋游戏(人 VS 人)
【C++ 程序】 井字棋游戏(人 VS Lv2电脑)
【C++ 程序】 井字棋游戏(人 VS Lv3电脑)
【C++ 程序】 井字棋游戏(人 VS Lv3电脑)(战绩统计版)
【C++ 程序】 五子棋游戏(人 VS 人)
【C++ 程序】 随机数
【C++ 程序】 移动迷宫游戏
【C++ 程序】 贪吃蛇游戏
【C++ 程序】 数字推盘游戏(15-puzzle)
【C++ 程序】 2048游戏
【C++ 程序】 井字棋游戏(人 VS 人)(EasyX 图形界面)
【C++ 程序】 井字棋游戏(人 VS Lv3电脑)(战绩统计版)(EasyX 图形界面)