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【C语言】三子棋实现AI智能落子（简单语法）_c语言落子算法

作者：不正经 | 2024-05-31 08:43:36

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c语言落子算法

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一、实现逻辑

  • 落子优先级（叙述为电脑视角）
    • 一手占中，若中心被占则占角（默认为电脑后手）
    • 有三连机会，尝试三连
    • 玩家有三连机会，进行拦截
    • 靠近己方棋子落子
  • 实现方法均为基础的循环和判断语句，并且适用于不同规格的棋盘

  • 基本逻辑：判断同行、同列、正副对角线是否有两棋相同，且可能三连的位置为空

    • 以同列且两棋不相邻为例（为确保一般性，这里使用4x4棋盘）。
    • 棋盘中实心红点为判断起始点（if语句的参考点）的所有可能位置，右侧为三连机会示例，中心为可能存在的制胜棋位置

     其余情况均为此思考模式，注意讨论对角线时需区分正副对角线

二、实现代码

1.一手占中，若中心被占则占角

代码如下：

  1. //电脑第一步
  2. void FirstMove(char board[ROW][COL], int row, int col)
  3. {
  4. 	int ar1[2] = { 0, row - 1 };
  5. 	int ar2[2] = { 0, col - 1 };
  6. 	int x = ar1[rand() % (row - 1)], y = ar2[rand() % (col - 1)];
  7. 	//判断中心点是否被占,玩家先中则不走边
  8. 	if (board[row / 2][col / 2] == '*')
  9. 	{
  10. 		while (1)
  11. 		{
  12. 			if (board[x][y] == ' ')
  13. 			{
  14. 				board[x][y] = '#';
  15. 				break;
  16. 			}
  17. 		}
  18. 	}
  19. 	//反则占中心
  20. 	else
  21. 		board[row / 2][col / 2] = '#';
  22. }

2.有三连机会，尝试三连

代码如下：

  1. //电脑三连机会，三连成功返回0，反之返回1
  2. int ComputerWin(char board[ROW][COL], int row, int col)
  3. {
  4. 	//1.同一行
  5. 	int i = 0, j = 0;
  6. 	//1.1一行相同两枚棋子相邻，判断参考点起点[0][0]，终点[row-1][col-2]
  7. 	for (i = 0; i < row; i++)
  8. 	{
  9. 		for (j = 0; j < col - 1; j++)
  10. 		{
  11. 			if (board[i][j] == board[i][j + 1] && board[i][j] == '#')
  12. 			{
  13. 				if (j < col - 2 && board[i][j + 2] == ' ')
  14. 				{
  15. 					board[i][j + 2] = '#';
  16. 					return 0;
  17. 				}
  18. 				else
  19. 					if (board[i][j - 1] == ' ')
  20. 					{
  21. 						board[i][j - 1] = '#';
  22. 						return 0;
  23.  
  24. 					}
  25. 			}
  26. 		}
  27. 	}
  28. 	//1.2一行相同两枚棋子不相邻，判断参考点起点[0][0]，终点[row-1][col-3]
  29. 	for (i = 0; i < row; i++)
  30. 	{
  31. 		for (j = 0; j < col - 2; j++)
  32. 		{
  33. 			if (board[i][j] == board[i][j + 2] && board[i][j] == '#' )
  34. 			{
  35. 				if (board[i][j + 1] == ' ')
  36. 				{
  37. 					board[i][j + 1] = '#';
  38. 					return 0;
  39. 				}
  40. 			}
  41. 		}
  42. 	}
  43. 	//2.同一列
  44. 	//2.1一列相同两枚棋子相邻，判断参考点起点[0][0]，终点[col - 2][row - 1]
  45. 	for (i = 0; i < row - 1; i++)
  46. 	{
  47. 		for (j = 0; j < col; j++)
  48. 		{
  49. 			if (board[i][j] == board[i + 1][j] && board[i][j] == '#')
  50. 			{
  51. 				if (i < col - 2 && board[i + 2][j] == ' ')
  52. 				{
  53. 					board[i + 2][j] = '#';
  54. 					return 0;
  55.  
  56. 				}
  57. 				else
  58. 					if (board[i - 1][j] == ' ')
  59. 					{
  60. 						board[i - 1][j] = '#';
  61. 						return 0;
  62.  
  63. 					}
  64. 			}
  65. 		}
  66. 	}
  67. 	//2.2一列相同两枚棋子不相邻，判断参考点起点[0][0]，终点[col - 3][row - 1]
  68. 	for (i = 0; i < row - 2; i++)
  69. 	{
  70. 		for (j = 0; j < col; j++)
  71. 		{
  72. 			if (board[i][j] == board[i + 2][j] && board[i][j] == '#' && board[i + 1][j] == ' ')
  73. 			{
  74. 				board[i + 1][j] = '#';
  75. 				return 0;
  76.  
  77. 			}
  78. 		}
  79. 	}
  80. 	//3.拦截对角线
  81. 	//3.1正对角线
  82. 	for (i = 0; i < row - 2; i++)
  83. 	{
  84. 		for (j = 0; j < col - 2; j++)
  85. 		{
  86. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '#')
  87. 			{
  88. 				if (board[i + 2][j + 2] == ' ')
  89. 				{
  90. 					board[i + 2][j + 2] = '#';
  91. 					return 0;
  92. 				}
  93. 			}
  94. 			if (board[i][j] == board[i + 2][j + 2] && board[i][j] == '#')
  95. 			{
  96. 				if (board[i + 1][j + 1] == ' ')
  97. 				{
  98. 					board[i + 1][j + 1] = '#';
  99. 					return 0;
  100. 				}
  101. 			}
  102. 		}
  103. 	}
  104. 	for (i = 1; i < row - 1; i++)
  105. 	{
  106. 		for (j = 1; j < col - 1; j++)
  107. 		{
  108. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '#')
  109. 			{
  110. 				if (board[i - 1][j - 1] == ' ')
  111. 				{
  112. 					board[i - 1][j - 1] = '#';
  113. 					return 0;
  114. 				}
  115.  
  116. 			}
  117. 		}
  118. 	}
  119. 	//3.2副对角线
  120. 	for (i = 2; i < row; i++)
  121. 	{
  122. 		for (j = 0; j < col - 1; j++)
  123. 		{
  124. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '#')
  125. 			{
  126. 				if (board[i - 2][j + 2] == ' ')
  127. 				{
  128. 					board[i - 2][j + 2] = '#';
  129. 					return 0;
  130. 				}
  131.  
  132. 			}
  133. 			if (board[i][j] == board[i - 2][j + 2] && board[i][j] == '#')
  134. 			{
  135. 				if (board[i - 1][j + 1] == ' ')
  136. 				{
  137. 					board[i - 1][j + 1] = '#';
  138. 					return 0;
  139. 				}
  140.  
  141. 			}
  142. 		}
  143. 	}
  144. 	for (i = 1; i < row - 1; i++)
  145. 	{
  146. 		for (j = 1; j < col - 1; j++)
  147. 		{
  148. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '#')
  149. 			{
  150. 				if (board[i + 1][j - 1] == ' ')
  151. 				{
  152. 					board[i + 1][j - 1] = '#';
  153. 					return 0;
  154. 				}
  155.  
  156. 			}
  157. 		}
  158. 	}
  159. 	return 1;
  160. }

3.玩家有三连机会，进行拦截

首先判断玩家是否存在三连机会，代码如下：

  1. //判断是否需要拦截，需要即返回1，反之返回0
  2. int IsDanger(char board[ROW][COL], int row, int col)
  3. {
  4. 	int i = 0, j = 0;
  5. 	//1.同一行三连
  6. 	//1.1一行相同两枚棋子相邻，判断参考点起点[0][0]，终点[row-1][col-2]
  7. 	for (i = 0; i < row; i++)
  8. 	{
  9. 		for (j = 0; j < col - 1; j++)
  10. 		{
  11. 			if (board[i][j] == board[i][j + 1] && board[i][j] == '*')
  12. 			{
  13. 				if (j < col - 2 && board[i][j + 2] == ' ')
  14. 				{
  15. 					return 1;
  16. 				}
  17. 				else
  18. 					if (board[i][j - 1] == ' ')
  19. 					{
  20. 						return 1;
  21. 					}
  22. 			}
  23. 		}
  24. 	}
  25. 	//1.2一行相同两枚棋子不相邻，判断参考点起点[0][0]，终点[row-1][col-3]
  26. 	for (i = 0; i < row; i++)
  27. 	{
  28. 		for (j = 0; j < col - 2; j++)
  29. 		{
  30. 			if (board[i][j] == board[i][j + 2] && board[i][j] == '*' && board[i][j + 1] == ' ')
  31. 			{
  32. 				return 1;
  33. 			}
  34. 		}
  35. 	}
  36. 	//2.同一列三连
  37. 	//2.1一列相同两枚棋子相邻，判断参考点起点[0][0]，终点[col - 2][row - 1]
  38. 	for (i = 0; i < row - 1; i++)
  39. 	{
  40. 		for (j = 0; j < col; j++)
  41. 		{
  42. 			if (board[i][j] == board[i + 1][j] && board[i][j] == '*')
  43. 			{
  44. 				if (i < col - 2 && board[i + 2][j] == ' ')
  45. 				{
  46. 					return 1;
  47. 				}
  48. 				else
  49. 					if (board[i - 1][j] == ' ')
  50. 					{
  51. 						return 1;
  52. 					}
  53. 			}
  54. 		}
  55. 	}
  56. 	//2.2一列相同两枚棋子不相邻，判断参考点起点[0][0]，终点[col - 3][row - 1]
  57. 	for (i = 0; i < row - 2; i++)
  58. 	{
  59. 		for (j = 0; j < col; j++)
  60. 		{
  61. 			if (board[i][j] == board[i + 2][j] && board[i][j] == '*' && board[i + 1][j] == ' ')
  62. 			{
  63. 				return 1;
  64. 			}
  65. 		}
  66. 	}
  67. 	//3.对角线
  68. 	//3.1正对角线
  69. 	for (i = 0; i < row - 2; i++)
  70. 	{
  71. 		for (j = 0; j < col - 2; j++)
  72. 		{
  73. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '*')
  74. 			{
  75. 				if (board[i + 2][j + 2] == ' ')
  76. 				{
  77. 					return 1;
  78. 				}
  79. 			}
  80. 			if (board[i][j] == board[i + 2][j + 2] && board[i][j] == '*')
  81. 			{
  82. 				if (board[i + 1][j + 1] == ' ')
  83. 				{
  84. 					return 1;
  85. 				}
  86.  
  87. 			}
  88. 		}
  89. 	}
  90. 	for (i = 1; i < row - 1; i++)
  91. 	{
  92. 		for (j = 1; j < col - 1; j++)
  93. 		{
  94. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '*')
  95. 			{
  96. 				if (board[i - 1][j - 1] == ' ')
  97. 				{
  98. 					return 1;
  99. 				}
  100.  
  101. 			}
  102. 		}
  103. 	}
  104. 	//3.2副对角线
  105. 	for (i = 2; i < row; i++)
  106. 	{
  107. 		for (j = 0; j < col - 1; j++)
  108. 		{
  109. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '*')
  110. 			{
  111. 				if (board[i - 2][j + 2] == ' ')
  112. 				{
  113. 					return 1;
  114. 				}
  115. 			}
  116. 			if (board[i][j] == board[i - 2][j + 2] && board[i][j] == '*')
  117. 			{
  118. 				if (board[i - 1][j + 1] == ' ')
  119. 				{
  120. 					return 1;
  121. 				}
  122. 			}
  123. 		}
  124. 	}
  125. 	for (i = 1; i < row - 1; i++)
  126. 	{
  127. 		for (j = 1; j < col - 1; j++)
  128. 		{
  129. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '*')
  130. 			{
  131. 				if (board[i + 1][j - 1] == ' ')
  132. 				{
  133. 					return 1;
  134. 				}
  135. 			}
  136. 		}
  137. 	}
  138. 	return 0;
  139. }

然后进行拦截，代码如下：

  1. //拦截玩家函数
  2. int BlockPlayer(char board[ROW][COL], int row, int col)
  3. {
  4. 	//1.同一行
  5. 	int i = 0, j = 0;
  6. 	//1.1一行相同两枚棋子相邻，判断参考点起点[0][0]，终点[row-1][col-2]
  7. 	for (i = 0; i < row; i++)
  8. 	{
  9. 		for (j = 0; j < col-1; j++)
  10. 		{
  11. 			if (board[i][j] == board[i][j + 1] && board[i][j] == '*')
  12. 			{
  13. 				if (j < col - 2 && board[i][j + 2] == ' ')
  14. 				{
  15. 					board[i][j + 2] = '#';
  16. 					return 0;
  17. 				}
  18. 				else
  19. 					if (board[i][j - 1] == ' ')
  20. 					{
  21. 						board[i][j - 1] = '#';
  22. 						return 0;
  23. 					}
  24. 			}
  25. 		}
  26. 	}
  27. 	//1.2一行相同两枚棋子不相邻，判断参考点起点[0][0]，终点[row-1][col-3]
  28. 	for (i = 0; i < row; i++)
  29. 	{
  30. 		for (j = 0; j < col - 2; j++)
  31. 		{
  32. 			if (board[i][j] == board[i][j + 2] && board[i][j] == '*' && board[i][j + 1] == ' ')
  33. 			{
  34. 				board[i][j + 1] = '#';
  35. 				return 0;
  36. 			}
  37. 		}
  38. 	}
  39. 	//2.同一列
  40. 	//2.1一列相同两枚棋子相邻，判断参考点起点[0][0]，终点[col - 2][row - 1]
  41. 	for (i = 0; i < row - 1; i++)
  42. 	{
  43. 		for (j = 0; j < col; j++)
  44. 		{
  45. 			if (board[i][j] == board[i + 1][j] && board[i][j] == '*')
  46. 			{
  47. 				if (i < col - 2 && board[i + 2][j] == ' ')
  48. 				{
  49. 					board[i + 2][j] = '#';
  50. 					return 0;
  51. 				}
  52. 				else
  53. 					if (board[i - 1][j] == ' ')
  54. 					{
  55. 						board[i - 1][j] = '#';
  56. 						return 0;
  57. 					}
  58. 			}
  59. 		}
  60. 	}
  61. 	//2.2一列相同两枚棋子不相邻，判断参考点起点[0][0]，终点[col - 3][row - 1]
  62. 	for (i = 0; i < row - 2; i++)
  63. 	{
  64. 		for (j = 0; j < col; j++)
  65. 		{
  66. 			if (board[i][j] == board[i + 2][j] && board[i][j] == '*' && board[i + 1][j] == ' ')
  67. 			{
  68. 				board[i + 1][j] = '#';
  69. 				return 0;
  70. 			}
  71. 		}
  72. 	}
  73. 	//3.对角线
  74. 	//3.1正对角线
  75. 	for (i = 0; i < row - 2; i++)
  76. 	{
  77. 		for (j = 0; j < col - 2; j++)
  78. 		{
  79. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '*')
  80. 			{
  81. 				if (board[i + 2][j + 2] == ' ')
  82. 				{
  83. 					board[i + 2][j + 2] = '#';
  84. 					return 0;
  85. 				}
  86. 			}
  87. 			if (board[i][j] == board[i + 2][j + 2] && board[i][j] == '*')
  88. 			{
  89. 				if (board[i + 1][j + 1] == ' ')
  90. 				{
  91. 					board[i + 1][j + 1] = '#';
  92. 					return 0;
  93. 				}
  94. 				
  95. 			}
  96. 		}
  97. 	}
  98. 	for (i = 1; i < row - 1; i++)
  99. 	{
  100. 		for (j = 1; j < col - 1; j++)
  101. 		{
  102. 			if (board[i][j] == board[i + 1][j + 1] && board[i][j] == '*')
  103. 			{
  104. 				if (board[i - 1][j - 1] == ' ')
  105. 				{
  106. 					board[i - 1][j - 1] = '#';
  107. 					return 0;
  108. 				}
  109. 				
  110. 			}
  111. 		}
  112. 	}
  113. 	//3.2副对角线
  114. 	for (i = 2; i < row; i++)
  115. 	{
  116. 		for (j = 0; j < col - 1; j++)
  117. 		{
  118. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '*')
  119. 			{
  120. 				if (board[i - 2][j + 2] == ' ')
  121. 				{
  122. 					board[i - 2][j + 2] = '#';
  123. 					return 0;
  124. 				}
  125. 			}
  126. 			if (board[i][j] == board[i - 2][j + 2] && board[i][j] == '*')
  127. 			{
  128. 				if (board[i - 1][j + 1] == ' ')
  129. 				{
  130. 					board[i - 1][j + 1] = '#';
  131. 					return 0;
  132. 				}
  133. 			}
  134. 		}
  135. 	}
  136. 	for (i = 1; i < row - 1; i++)
  137. 	{
  138. 		for (j = 1; j < col - 1; j++)
  139. 		{
  140. 			if (board[i][j] == board[i - 1][j + 1] && board[i][j] == '*')
  141. 			{
  142. 				if (board[i + 1][j - 1] == ' ')
  143. 				{
  144. 					board[i + 1][j - 1] = '#';
  145. 					return 0;
  146. 				}
  147. 			}
  148. 		}
  149. 	}
  150. 	return 1;
  151. }

4.无三连机会，且无需拦截，则靠近己方棋子落子，创造三连机会

代码如下：

  1. //无三连机会，无需拦截，自行制造机会
  2. int CreatWin(char board[ROW][COL], int row, int col)
  3. {
  4. 	int i = 0, j = 0;
  5. 	int a = 0, b = 0;
  6. 	for (i = 0; i < row; i++)
  7. 	{
  8. 		for (j = 0; j < col; j++)
  9. 		{
  10. 			if (board[i][j] == '#')
  11. 			{
  12. 				for (a = 0; a < row; a++)
  13. 				{
  14. 					for (b = 0; b < col; b++)
  15. 					{
  16. 						if ((a - i == -1 && b - j == -1)
  17. 							|| (a - i == -1 && b - j == 0)
  18. 							|| (a - i == -1 && b - j == 1)
  19. 							|| (a - i == 0 && b - j == -1)
  20. 							|| (a - i == 0 && b - j == 1)
  21. 							|| (a - i == 1 && b - j == -1)
  22. 							|| (a - i == 1 && b - j == 0)
  23. 							|| (a - i == 1 && b - j == 1))
  24. 						{
  25. 							if (board[a][b] == ' ')
  26. 							{
  27. 								board[a][b] = '#';
  28. 								return 0;
  29. 							}
  30. 						}
  31. 					}
  32. 				}
  33. 			}
  34. 		}
  35. 	}
  36. 	
  37. }

总结

1-3步均使用一套判断模式，第4步可应用在扫雷中的排查雷的步骤中。

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