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机器博弈中的数据结构与基本方法（三）-----博弈树的搜索之九宫重排问题

　　九宫重排问题，如下图所示，九宫重排（或称八数码问题）是人工智能中一个经典问题，这里利用DFS解决这个问题。

　　

 　　通过设置一个OPEN表和一个CLOSE表记录尚未遍历和遍历过的状态，直至找到目标状态或遍历完所有状态，如果遍历完所有状态还木有找到目标状态则无解。

　　代码

  1 #include <cstdio>
  2 //利用深度优先搜索
  3 //构造OPEN表和CLOSE表
  4 typedef struct {
  5     int State[5][5];
  6     int FatherState[5][5];
  7 }OPEN;
  8 typedef struct {
  9     int Order;
 10     int State[5][5];
 11     int FatherState[5][5];
 12 }CLOSE;
 13 OPEN OpenList[500000];
 14 CLOSE CloseList[500000];
 15 int OpenCount = -1, CloseCount = -1;
 16 //判断位置是否合法的辅助数组
 17 int Position[5][5] = {
 18     {-1, -1, -1, -1, -1},
 19     {-1,  1,  1,  1, -1},
 20     {-1,  1,  1,  1, -1},
 21     {-1,  1,  1,  1, -1},
 22     {-1, -1, -1, -1, -1}
 23 };
 24 //将某一状态存入OPEN表
 25 void Push(int a[][5], int b[][5]){
 26     OpenCount++;
 27     for (int i = 0; i < 5; i++){
 28         for (int j = 0; j < 5; j++){
 29             OpenList[OpenCount].State[i][j] = a[i][j];
 30             OpenList[OpenCount].FatherState[i][j] = b[i][j];
 31         }
 32     }
 33 }
 34 //比较连个状态是否完全相同
 35 int Compare(int a[][5], int b[][5]){
 36     for (int i = 0; i < 5; i++){
 37         for (int j = 0; j < 5; j++){
 38             if (a[i][j] != b[i][j]){
 39                 return 0;
 40             }
 41         }
 42     } 
 43     return 1;
 44 }
 45 
 46 void Move(int a[][5]){
 47     int temp[5][5];
 48     int i, j, m, n, flag1 = 0, flag2;
 49     //找到0的位置
 50     for (i = 0; i < 5; i++){
 51         for (j = 0; j < 5; j++){
 52             if (0 == a[i][j]){
 53                 flag1 = 1;
 54                 break;
 55             }
 56         } 
 57         if (flag1){
 58             break;
 59         }
 60     }
 61     //复制数组
 62     for (m = 0; m < 5; m++){
 63         for (n = 0; n < 5; n++){
 64             temp[m][n] = a[m][n];
 65         }
 66     }
 67     if (-1 != Position[i][j-1]){                 //方向 1
 68         a[i][j] = a[i][j-1];
 69         a[i][j-1] = 0;                           //移动数字
 70         flag2 = 0;
 71         for (int k = 0; k <= OpenCount; k++){     //检查当前状态是否在OPEN表中
 72             if (Compare(OpenList[k].State, a)){
 73                 flag2 = 1;
 74                 break;
 75             }
 76         }
 77         if (!flag2){
 78             for (int k = 0; k <= CloseCount; k++){//检查当前状态是否在CLOSE表中
 79                 if (Compare(CloseList[k].State, a)){
 80                     flag2 = 1;
 81                     break;
 82                 }
 83             }
 84         }
 85         if (!flag2){                                   //若果都不在就加入OPEN表
 86             Push(a, temp);
 87         }
 88         for (m = 0; m < 5; m++){                    //复原数组a,做下一移动方向的判断
 89             for (n = 0; n < 5; n++){
 90                 a[m][n] = temp[m][n];
 91             }
 92         }
 93     }
 94 
 95 
 96     if (-1 != Position[i][j+1]){                 //方向 2
 97         a[i][j] = a[i][j+1];
 98         a[i][j+1] = 0;                           //移动数字
 99         flag2 = 0;
100         for (int k = 0; k <= OpenCount; k++){
101             if (Compare(OpenList[k].State, a)){
102                 flag2 = 1;
103                 break;
104             }
105         }
106         if (!flag2){
107             for (int k = 0; k <= CloseCount; k++){
108                 if (Compare(CloseList[k].State, a)){
109                     flag2 = 1;
110                     break;
111                 }
112             }
113         }
114         if (!flag2){
115             Push(a, temp);
116         }
117         for (m = 0; m < 5; m++){
118             for (n = 0; n < 5; n++){
119                 a[m][n] = temp[m][n];
120             }
121         }
122     }
123 
124     if (-1 != Position[i+1][j]){                 //方向 3
125         a[i][j] = a[i+1][j];
126         a[i+1][j] = 0;                           //移动数字
127         flag2 = 0;
128         for (int k = 0; k <= OpenCount; k++){
129             if (Compare(OpenList[k].State, a)){
130                 flag2 = 1;
131                 break;
132             }
133         }
134         if (!flag2){
135             for (int k = 0; k <= CloseCount; k++){
136                 if (Compare(CloseList[k].State, a)){
137                     flag2 = 1;
138                     break;
139                 }
140             }
141         }
142         if (!flag2){
143             Push(a, temp);
144         }
145         for (m = 0; m < 5; m++){
146             for (n = 0; n < 5; n++){
147                 a[m][n] = temp[m][n];
148             }
149         }
150     }
151 
152     if (-1 != Position[i-1][j]){                 //方向 4
153         a[i][j] = a[i-1][j];
154         a[i-1][j] = 0;                           //移动数字
155         flag2 = 0;
156         for (int k = 0; k <= OpenCount; k++){
157             if (Compare(OpenList[k].State, a)){
158                 flag2 = 1;
159                 break;
160             }
161         }
162         if (!flag2){
163             for (int k = 0; k <= CloseCount; k++){
164                 if (Compare(CloseList[k].State, a)){
165                     flag2 = 1;
166                     break;
167                 }
168             }
169         }
170         if (!flag2){
171             Push(a, temp);
172         }
173         for (m = 0; m < 5; m++){
174             for (n = 0; n < 5; n++){
175                 a[m][n] = temp[m][n];
176             }
177         }
178     }
179 }
180 
181 //主函数
182 int main(int argc, char const *argv[])
183 {
184     //输出位置
185     freopen("ans.txt", "w", stdout);
186     //起始状态
187     int a[5][5], w[5][5]={
188         {-1, -1, -1, -1, -1},
189         {-1,  1,  3,  4, -1},
190         {-1,  8,  2,  5, -1},
191         {-1,  7,  0,  6, -1},
192         {-1, -1, -1, -1, -1}
193     };
194     //目标状态
195     int result[5][5] = {
196         {-1, -1, -1, -1, -1},
197         {-1,  1,  2,  3, -1},
198         {-1,  8,  0,  4, -1},
199         {-1,  7,  6,  5, -1},
200         {-1, -1, -1, -1, -1}
201     };
202 
203     Push(w, w);
204     while (-1 != OpenCount){//OPEN表不为空
205         CloseCount++;
206         //每完成一次移动，该状态即可放入CLOSE表中
207         for (int i = 0; i < 5; i++){
208             for (int j = 0; j < 5; j++){
209                 CloseList[CloseCount].State[i][j] = OpenList[OpenCount].State[i][j];                
210                 CloseList[CloseCount].FatherState[i][j] = OpenList[OpenCount].FatherState[i][j];                
211             }
212         }
213         CloseList[CloseCount].Order = CloseCount;
214         //更新当前状态
215         for (int i = 0; i < 5; i++){
216             for (int j = 0; j < 5; j++){
217                 a[i][j] = OpenList[OpenCount].State[i][j];                
218             }
219         }
220         OpenCount--;
221         if (Compare(result, a)){    //如果达到了目标状态则技术循环
222             break;
223         } else {
224             Move(a);                //继续移动
225         }
226     }
227     //输出到达目标状态的着法
228     for (int k = 0; k <= CloseCount; k++){
229         printf("No: %d\n", CloseList[k].Order);
230         printf("  Parent\t\tChild\n");
231         for (int i = 1; i < 4; i++){
232             for (int j = 1; j < 4; j++){
233                 printf("%3d", CloseList[k].FatherState[i][j]);                           
234             }
235             printf("\t");
236             for (int j = 1; j < 4; j++){
237                 printf("%3d", CloseList[k].State[i][j]);               
238             }
239             printf("\n");
240         }
241     }
242     return 0;
243 }
View Code

　　结果

No: 0
  Parent        Child
  1  3  4      1  3  4
  8  2  5      8  2  5
  7  0  6      7  0  6
No: 1
  Parent        Child
  1  3  4      1  3  4
  8  2  5      8  0  5
  7  0  6      7  2  6
No: 2
  Parent        Child
  1  3  4      1  0  4
  8  0  5      8  3  5
  7  2  6      7  2  6
No: 3
  Parent        Child
  1  0  4      1  4  0
  8  3  5      8  3  5
  7  2  6      7  2  6
No: 4
  Parent        Child
  1  4  0      1  4  5
  8  3  5      8  3  0
  7  2  6      7  2  6
No: 5
  Parent        Child
  1  4  5      1  4  5
  8  3  0      8  3  6
  7  2  6      7  2  0
No: 6
  Parent        Child
  1  4  5      1  4  5
  8  3  6      8  3  6
  7  2  0      7  0  2
No: 7
  Parent        Child
  1  4  5      1  4  5
  8  3  6      8  0  6
  7  0  2      7  3  2
No: 8
  Parent        Child
  1  4  5      1  0  5
  8  0  6      8  4  6
  7  3  2      7  3  2
No: 9
  Parent        Child
  1  0  5      1  5  0
  8  4  6      8  4  6
  7  3  2      7  3  2
No: 10
  Parent        Child
  1  5  0      1  5  6
  8  4  6      8  4  0
  7  3  2      7  3  2
No: 11
  Parent        Child
  1  5  6      1  5  6
  8  4  0      8  4  2
  7  3  2      7  3  0
No: 12
  Parent        Child
  1  5  6      1  5  6
  8  4  2      8  4  2
  7  3  0      7  0  3
No: 13
  Parent        Child
  1  5  6      1  5  6
  8  4  2      8  0  2
  7  0  3      7  4  3
No: 14
  Parent        Child
  1  5  6      1  0  6
  8  0  2      8  5  2
  7  4  3      7  4  3
No: 15
  Parent        Child
  1  0  6      1  6  0
  8  5  2      8  5  2
  7  4  3      7  4  3
No: 16
  Parent        Child
  1  6  0      1  6  2
  8  5  2      8  5  0
  7  4  3      7  4  3
No: 17
  Parent        Child
  1  6  2      1  6  2
  8  5  0      8  5  3
  7  4  3      7  4  0
No: 18
  Parent        Child
  1  6  2      1  6  2
  8  5  3      8  5  3
  7  4  0      7  0  4
No: 19
  Parent        Child
  1  6  2      1  6  2
  8  5  3      8  0  3
  7  0  4      7  5  4
No: 20
  Parent        Child
  1  6  2      1  0  2
  8  0  3      8  6  3
  7  5  4      7  5  4
No: 21
  Parent        Child
  1  0  2      1  2  0
  8  6  3      8  6  3
  7  5  4      7  5  4
No: 22
  Parent        Child
  1  2  0      1  2  3
  8  6  3      8  6  0
  7  5  4      7  5  4
No: 23
  Parent        Child
  1  2  3      1  2  3
  8  6  0      8  6  4
  7  5  4      7  5  0
No: 24
  Parent        Child
  1  2  3      1  2  3
  8  6  4      8  6  4
  7  5  0      7  0  5
No: 25
  Parent        Child
  1  2  3      1  2  3
  8  6  4      8  0  4
  7  0  5      7  6  5
View Code

　　备注

　　这个八数码问题可以做成一个类似2048的游戏，不过挑战难度显然更高，不过现在GUI正在学习，恐怕要过一段时间才能做出demo。

　　本书最后一个是关于MAX-MIN启发式搜索，但是看的不是很明白，也从来没用过，等到以后需要再来略作研究。

posted @ 2016-07-04 19:58  赵裕(vimerzhao)  阅读(639)  评论(0)  编辑  收藏  举报