标签：image 过程 ret 策略 ESS switch bre 代码 png
游戏介绍
井字棋，英文名叫Tic-Tac-Toe，是一种在3*3格子上进行的连珠游戏，和五子棋类似。然后由分别代表O和X的两个游戏者轮流在格子里留下标记（一般来说先手者为X），任意三个标记形成一条直线（包括行、列、对角线、反对角线），则为获胜。
解决策略
重点在于电脑方如何下棋，我们采取估计棋局每个位置的权重，首先要对棋局进行分类。
---3个为空，重要性最低，权值设置为1 //视为暂时不管
---2个空1个对方，重要性次低，权值为10 //一下三个区别不大，可比较随意的设置
----1个空格1个对方1个己方，重要行较低，权值50
----2个空格1个己方，重要性较高，权值为100
---1个空格2个对方，重要性次高，权值500 //不堵住会输
---1个空格2个己方，重要性最高，权值1000 //可以直接赢
注意几点：
1、权值之间的间距可以设大一点
2、对每个空位置，权值等于行权值+列权值+对角线权值+反对角线权值，这4中权值都可以用上面的估算，但不做改进会出bug
考虑如下情况：
(1,3)-->(3,1)-->(1,1)-->(2,1)电脑就输了
---->
--->
-->
-->人获胜
关键在于第二步，应该选择一个非角的位置，原因在于此时右上角位置的权值大于中上位置，分析权值的来源
右上角时，10+10+100(对角线己方），而中上时，10+100(行己方)，所以同样是2空1己方时，己方位于行或列的权重应大于己方位于对角线。
所以按行或列计算时，2空1己方的权值可改为200
代码实现
1 #include<stdio.h> 2 #include<Windows.h> 3 4 const int ROW = 3; 5 const int COL = 3; 6 int chessboard[ROW][COL]; 7 int score[ROW][COL]; 8 9 void Initmap(); 10 void Showmap(); //打印棋局 11 bool isWin(); //判断是否有一方获胜 12 bool isFull(); //判断棋盘是否为满 13 void PcPlay(); //电脑下棋 14 void HumanPlay(); //人下棋 15 16 int main() 17 { 18 Initmap(); 19 Showmap(); 20 while ((!isFull()) && (!isWin())) 21 { 22 HumanPlay(); 23 system("cls"); 24 Showmap(); 25 if (isWin()) 26 break; 27 28 Sleep(500); //模拟实际过程，让电脑慢点，hh 29 PcPlay(); 30 system("cls"); 31 Showmap(); 32 } 33 34 if (isFull()) 35 printf("\n\n平局\n"); 36 37 system("pause"); 38 return 0; 39 } 40 41 void Initmap() 42 { 43 for (int i = 0; i < ROW; i++) 44 for (int j = 0; j < COL; j++) 45 chessboard[i][j] = 1; 46 } 47 48 void Showmap() 49 { 50 for (int i = 0; i < ROW; i++) 51 { 52 for (int j = 0; j < COL; j++) 53 { 54 if (chessboard[i][j] == 1) //"1"代表空 55 printf("□"); 56 if (chessboard[i][j] == 2) //"2"代表人 57 printf("■"); 58 if (chessboard[i][j] == 5) //"5"代表电脑 59 printf("●"); 60 } 61 printf("\n"); 62 } 63 } 64 65 bool isWin() 66 { 67 int sum = 0; 68 for (int i = 0; i < ROW; i++) //对每行判断是否获胜 69 { 70 for (int j = 0; j < COL; j++) 71 sum += chessboard[i][j]; 72 73 if (sum == 6) 74 { 75 printf("人获胜！\n"); 76 return true; 77 } 78 if (sum == 15) 79 { 80 printf("电脑获胜！\n"); 81 return true; 82 } 83 sum = 0; 84 } 85 86 for (int j = 0; j < ROW; j++) //对每列判断是否获胜 87 { 88 for (int i = 0; i < COL; i++) 89 sum += chessboard[i][j]; 90 91 if (sum == 6) 92 { 93 printf("人获胜！\n"); 94 return true; 95 } 96 if (sum == 15) 97 { 98 printf("电脑获胜！\n"); 99 return true; 100 } 101 sum = 0; 102 } 103 104 for (int i = 0; i < ROW; i++) //对对角线判断是否获胜 105 sum += chessboard[i][i]; 106 if (sum == 6) 107 { 108 printf("人获胜！\n"); 109 return true; 110 } 111 if (sum == 15) 112 { 113 printf("电脑获胜！\n"); 114 return true; 115 } 116 117 sum = 0; 118 for (int i = 0; i < ROW; i++) //对反对角线判断是否获胜 119 sum += chessboard[i][2 - i]; 120 if (sum == 6) 121 { 122 printf("人获胜！\n"); 123 return true; 124 } 125 if (sum == 15) 126 { 127 printf("电脑获胜！\n"); 128 return true; 129 } 130 131 return false; 132 } 133 134 bool isFull() 135 { 136 for (int i = 0; i < ROW; i++) 137 for (int j = 0; j < COL; j++) 138 if (chessboard[i][j] == 1) 139 return false; 140 return true; 141 } 142 143 void HumanPlay() 144 { 145 int x, y; 146 printf("请输入棋子的横坐标X:"); 147 scanf_s("%d", &x); 148 printf("请输入棋子的纵坐标Y:"); 149 scanf_s("%d", &y); 150 151 while (x < 1 || x>3 || y < 1 || y>3) 152 { 153 printf("\n请正确输入！\n"); 154 printf("x,y均属于1~3\n\n"); 155 156 printf("请输入棋子的横坐标X:"); 157 scanf_s("%d", &x); 158 printf("请输入棋子的纵坐标Y:"); 159 scanf_s("%d", &y); 160 } 161 162 while (chessboard[3 - y][x - 1] != 1) 163 { 164 printf("\n\n该位置已被占用！\n"); 165 printf("请选择正确的位置\n\n"); 166 Sleep(1000); 167 168 printf("\n请输入棋子的横坐标X:"); 169 scanf_s("%d", &x); 170 printf("请输入棋子的纵坐标Y:"); 171 scanf_s("%d", &y); 172 } 173 174 chessboard[3 - y][x - 1] = 2; 175 } 176 177 void PcPlay() 178 { 179 int sum = 0; 180 for (int i = 0; i < ROW; i++) 181 for (int j = 0; j < COL; j++) 182 score[i][j] = 0; 183 184 // 对每行进行分数统计 185 for (int i = 0; i < ROW; i++) 186 { 187 for (int j = 0; j < COL; j++) 188 sum += chessboard[i][j]; 189 190 switch (sum) 191 { 192 case 3: //1+1+1;重要性：最低;权重：1 193 for (int k = 0; k < COL; k++) 194 { 195 if (chessboard[i][k] == 1) 196 score[i][k] += 1; 197 } 198 break; 199 case 4: //1+1+2;重要性：次低;权重：10 200 for (int k = 0; k < COL; k++) 201 { 202 if (chessboard[i][k] == 1) 203 score[i][k] += 10; 204 } 205 break; 206 case 8: //1+2+5;重要性：较低，权值50 207 for (int k = 0; k < COL; k++) 208 { 209 if (chessboard[i][k] == 1) 210 score[i][k] += 50; 211 } 212 break; 213 case 7: //1+1+5;重要性：较高;权重：200 214 for (int k = 0; k < COL; k++) 215 { 216 if (chessboard[i][k] == 1) 217 score[i][k] += 200; //把行列的重要性比对角线高 218 } 219 break; 220 case 5: //1+2+2;重要性：次高;权重：500 221 for (int k = 0; k < COL; k++) 222 { 223 if (chessboard[i][k] == 1) 224 score[i][k] += 500; 225 } 226 break; 227 case 11: //1+5+5;重要性：最高;权重：1000 228 for (int k = 0; k < COL; k++) 229 { 230 if (chessboard[i][k] == 1) 231 score[i][k] += 1000; 232 } 233 break; 234 } 235 sum = 0; 236 } 237 238 // 对每列进行分数统计 239 for (int j = 0; j < COL; j++) 240 { 241 for (int i = 0; i < ROW; i++) 242 sum += chessboard[i][j]; 243 244 switch (sum) 245 { 246 case 3: 247 for (int k = 0; k < COL; k++) 248 { 249 if (chessboard[k][j] == 1) 250 score[k][j] += 1; 251 } 252 break; 253 case 4: 254 for (int k = 0; k < COL; k++) 255 { 256 if (chessboard[k][j] == 1) 257 score[k][j] += 10; 258 } 259 break; 260 case 8: 261 for (int k = 0; k < 262 COL; k++) 263 { 264 if (chessboard[k][j] == 1) 265 score[k][j] += 50; 266 } 267 break; 268 case 7: 269 for (int k = 0; k < COL; k++) 270 { 271 if (chessboard[k][j] == 1) //1+1+5;重要性：较高;权重：200 272 score[k][j] += 200; 273 } 274 break; 275 case 5: 276 for (int k = 0; k < COL; k++) 277 { 278 if (chessboard[k][j] == 1) 279 score[k][j] += 500; 280 } 281 break; 282 case 11: 283 for (int k = 0; k < COL; k++) 284 { 285 if (chessboard[k][j] == 1) 286 score[k][j] += 1000; 287 } 288 break; 289 } 290 sum = 0; 291 } 292 293 // 对对角线进行分数统计 294 for (int i = 0; i < ROW; i++) 295 sum += chessboard[i][i]; 296 switch (sum) 297 { 298 case 3: 299 for (int i = 0; i < COL; i++) 300 { 301 if (chessboard[i][i] == 1) 302 score[i][i] += 1; 303 } 304 break; 305 case 4: 306 for (int i = 0; i < COL; i++) 307 { 308 if (chessboard[i][i] == 1) 309 score[i][i] += 10; 310 } 311 break; 312 case 8: 313 for (int i = 0; i < COL; i++) 314 { 315 if (chessboard[i][i] == 1) 316 score[i][i] += 50; 317 } 318 break; 319 case 7: //1+1+5;权重：100 320 for (int i = 0; i < COL; i++) 321 { 322 if (chessboard[i][i] == 1) 323 score[i][i] += 100; 324 } 325 break; 326 case 5: 327 for (int i = 0; i < COL; i++) 328 { 329 if (chessboard[i][i] == 1) 330 score[i][i] += 500; 331 } 332 break; 333 case 11: 334 for (int i = 0; i < COL; i++) 335 { 336 if (chessboard[i][i] == 1) 337 score[i][i] += 1000; 338 } 339 break; 340 } 341 342 // 对反对角线进行分数统计 343 sum = 0; 344 for (int i = 0; i < ROW; i++) 345 sum += chessboard[i][2 - i]; 346 switch (sum) 347 { 348 case 3: 349 for (int i = 0; i < COL; i++) 350 { 351 if (chessboard[i][2 - i] == 1) 352 score[i][2 - i] += 1; 353 } 354 break; 355 case 4: 356 for (int i = 0; i < COL; i++) 357 { 358 if (chessboard[i][2 - i] == 1) 359 score[i][2 - i] += 10; 360 } 361 break; 362 case 8: 363 for (int i = 0; i < COL; i++) 364 { 365 if (chessboard[i][2 - i] == 1) 366 score[i][2 - i] += 50; 367 } 368 break; 369 case 7: 370 for (int i = 0; i < COL; i++) 371 { 372 if (chessboard[i][2 - i] == 1) //1+1+5;权重：100 373 score[i][2 - i] += 100; 374 } 375 break; 376 case 5: 377 for (int i = 0; i < COL; i++) 378 { 379 if (chessboard[i][2 - i] == 1) 380 score[i][2 - i] += 500; 381 } 382 break; 383 case 11: 384 for (int i = 0; i < COL; i++) 385 { 386 if (chessboard[i][2 - i] == 1) 387 score[i][2 - i] += 1000; 388 } 389 break; 390 } 391 392 int maxRow = 0, maxCol = 0; 393 for (int i = 0; i < ROW; i++) 394 for (int j = 0; j < COL; j++) 395 { 396 if (score[i][j] > score[maxRow][maxCol]) 397 { 398 maxRow = i; 399 maxCol = j; 400 } 401 } 402 chessboard[maxRow][maxCol] = 5; 403 }
标签：image 过程 ret 策略 ESS switch bre 代码 png
原文地址：https://www.cnblogs.com/lfri/p/9880328.html
