A*算法实现猎人渡河问题 





  1 问题描述：
  2  猎人、狗、男人带着男孩1、男孩2，女人带着女孩1、女孩2，一共8个体。一艘小船，一次只能过两个个体，狗和四个小孩不会划船。
  3  注意：
  4 （1）猎人不在，狗咬任何人；
  5 （2）男人不在，女人打男孩；
  6 （3）女人不在，男人打女孩
  7 /*思路：用0表示在此岸，1表示在对岸(0,0,0,0,0,0,0,0,0)分别表示猎人，男人
  8 女人，狗狗，女孩，女孩，男孩，男孩，船在此岸。在满足条件是进行状态转移确定
  9 状态结点信息*/
 10 #include<iostream>
 11 #include<stdlib.h>
 12 #include"A_Def.h"
 13 #include"A_Algo.h"
 14 int main() {
 15     system("Color B5");
 16     int begin[9] = {0,0,0,0,0,0,0,0,0};
 17     int end[9] = {1,1,1,1,1,1,1,1,1};
 18     A_Star(begin,end);
 19     return 0;
 20 }
 21 using namespace std;
 22 void create(Astar &Head) {//创建一个初始化链表
 23     Head = (Astar)(malloc(sizeof(AS)));
 24     Head->next = NULL;
 25 }
 26 int Is_Empety(Astar Head) {//判断表是否为空
 27     if (Head->next!=NULL) return 1;
 28     else return 0;
 29 }
 30 int Is_SomeBefor(Astar Sstar, Astar Head) {//判断新生成的结点是否与其
 31 //上一状态的祖辈有相同的状态
 32     int i;
 33     while (Head!=NULL) {
 34         for (i = 0; i <= 8; i++) {
 35             if (Sstar->t[i] != Head->t[i])
 36                 break;
 37         }
 38         if (i == 9) return 1;
 39         Head = Head->parents;
 40     }return 0;
 41 }
 42 int Is_OK(Astar A) {//判断是否满足转移条件
 43     int F = 1;
 44     //if (A->t[0] != A->t[3]) F = 0;
 45     if((A->t[0]&&!A->t[3]&&(!A->t[1]||!A->t[2]||!A->t[4]||!A->t[5]||!A- >t[6]||!A->t[7]))||(!A->t[0]&&A->t[3]&&(A->t[1]||A->t[2]||A->t[4]||
 46     A->t[5]||A->t[6]||A->t[7]))) 
 48         F=0;
 49     if((A->t[1] && !A->t[2] && (!A->t[6] || !A->t[7]))|| (!A->t[1]&& A->t[2] && (A->t[6] || A->t[7])))
 51         F=0;
 52     if((A->t[2] && !A->t[1] && (!A->t[4] || !A->t[5])) || (!A->t[2] && A->t[1] (A->t[4] || A->t[5])))
 54         F=0;
 55     return F;
 56 }
 57 void Add_Open(Astar &Open,Astar &Check) {//将f排序加入
 58     Astar p, q;
 59     p = Open;
 60     q = Open->next;//
 61     while (q!=NULL&&q->f<Check->f) {
 62         p = q; 57
 63     Check->next = p->next;
 64     p->next = Check;
 65 }
 66 int G_value(Astar Sstar, Astar& Head) {//计算初始结点到当前考察的实际代价
 67     if (Head==NULL)  Sstar->g = 0;
 68     else {
 69         Sstar->g = Head->g + 1;
 70     }
 71     return Sstar->g;
 72 }
 73 int H_value(Astar Sstar) {//计算当前结点到目标结点的yuce值 核心2
 74     int sum = 0;
 75     if (!Sstar->t[8]) {
 76         for (int i = 0; i <= 7; i++) {
 77             if (!Sstar->t[i])
 78                 sum++;
 79         }
 80         if (sum >= 2) {
 81             Sstar->h = (sum - 2) * 2 + 1;
 82             return Sstar->h;
 83         }
 84         else { Sstar->h = 1; return 1; }
 85     }
 86     else 
 87     {
 88         for (int i = 0; i <= 7; i++) {
 89             if (!Sstar->t[i])
 90                 sum++;
 91         }
 92         
 93         Sstar->h = 2*sum;
 94         return Sstar->h;
 95     }
 96 }
 97 void Open_EnterFirst(int begin[],int end[]) {//Open表加入第一个结点
 98     Astar Open_First=Astar(malloc(sizeof(AS)));
 99     Open_First->parents = NULL;
100     Open_First->chilren = NULL;
101     Astar Q = NULL;
102     for (int i = 0; i <= 8; i++) Open_First->t[i] = begin[i];
103     Open_First->next = NULL;
104     Open_First->g = G_value(Open_First, Q); 
105     Open_First->h = H_value(Open_First);
106     Open_First->f = Open_First->g + Open_First->h; 
107     Add_Open(Open, Open_First);
108 }
109 void Out(Astar Head,Astar& H) {//从表中取出结点类似于尾插法取出
110     if (!Is_Empety(Head)) {
111         H = NULL;
112         return;
113     }
114     H = Head->next;
115     Head->next = Head->next->next;
116     H->next = NULL;
117 }
118 void Enter_Close(Astar Sstar, Astar& Check_L) {//待考察结点进入Close中或普通结点进普通表
119     Sstar->next = Check_L->next;
120     Check_L->next = Sstar;
121 }
122 void Out_Path(Astar A_Check) {//输出最快渡河的方案lujin
123     cout << "渡河代价:" << A_Check->f << endl;
124     //Astar A_Check1 = A_Check->next; 
125     int i = 0;
126     while (A_Check) {
127         if (A_Check->parents) {//确保运行成功
128             for (i = 0; i <= 8; i++) {
129                 cout << A_Check->t[i] << " ";
130                 if (i == 8) {
131                     int p = 1;
132                     if (!A_Check->parents->t[8]) {
133                         for (int q = 0; q <= 7; q++) {
134                             if (!A_Check->parents->t[q]) {//至少为0才过去
135                                 if (A_Check->t[q] != A_Check->parents->t[q]) {//与下一个状态不相等才过去
136                                     if (p == 1) cout << "<----";
137                                     if (q == 0) cout << "猎人" << "过河" <<" " ;
138                                     if (q == 1) cout << "男人" << "过河" << " ";
139                                     if (q == 2) cout << "女人" << "过河" << " ";
140                                     if (q == 3) cout << "狗狗" << "过河" << " ";
141                                     if (q == 4) cout << "女孩" << "过河" << " ";
142                                     if (q == 5) cout << "女孩" << "过河" << " ";
143                                     if (q == 6) cout << "男孩" << "过河" << " ";
144                                     if (q == 7) cout << "男孩" << "过河" << " ";
145                                     p++;
146                                 }
147                                 else continue;
148                             }
149                             else continue;
150                         }
151                     }
152                     else
153                     {
154                         int r = 1;
155                         for (int q = 0; q <= 7; q++) {
156                             if (A_Check->parents->t[q]) {//至少为1才过去
157                                 
158                                 if (A_Check->t[q] != A_Check->parents->t[q]) {//与下一个状态不相等才过去
159                                     if (r == 1) cout << "<----";
160                                     if (q == 0) cout << "猎人" << "回去" << " "; 
161                                     if (q == 1) cout << "男人" << "回去" << " ";
162                                     if (q == 2) cout << "女人" << "回去" << " ";
163                                     if (q == 3) cout << "狗狗" << "回去" << " ";
164                                     r++;
165                                 }
166                                 else continue;
167                             }
168                             else continue;
169                         }
170                     }
171                 }
172             }
173             cout << endl;
174             A_Check = A_Check->parents;
175         }
176     }
177 }
178 int Is_In(Astar  A_Check_L_One,Astar Check,Astar &M,Astar &N) {//判断某结点状态在哪个表中
179     int i; 
180     N = Check;
181     Check = Check->next;
182     while (Check !=NULL) {
183         for ( i = 0; i <= 8; i++) {  
184             if (A_Check_L_One->t[i] != Check->t[i])
185                 break;
186         }
187         if (i == 9) 
188         {
189             M = Check;
190             return 1; 
191         }
192         N = Check;
193         Check = Check->next;
194     }
195     return 0;
196 }
197 int Is_Same(Astar A, Astar Sstar) {//同一层是否重复
198     int i = 0; 
199     A = A->next; 
200     while (A) {
201         for (i = 0; i <= 8; i++) {
202             //cout << A->t[i] << "==" << Sstar->t[i] ;
203             if (A->t[i] != Sstar->t[i])
204                 break;
205         }
206         if (i == 9) return 1;
207         A = A->next;
208     }
209     return 0;
210 }
211 void A_Check_LS(Astar A_Check_L, Astar A_Check) {//通过将上一个状态传过来来实现状态转移并且确定状态结点信息(g,h,f),状态转移设置两种方式-->2*2=4//细节较多
212     //Astar Sstar = (Astar)malloc(sizeof(AS));
213     //Astar Sstar1 = (Astar)malloc(sizeof(AS));
214     int t0 ; 
215     //for (int i = 0; i <= 8; i++) cout << A_Check->t[i] << "++";; cout << endl;
216     int t1[9], t2[9];
217     for (int i = 0; i <= 8; i++)  t2[i] = A_Check->t[i];
218     if (!A_Check->t[8]) {    t0 = 1;
219       for (int i = 0; i <= 2; i++) {
220         for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];//操作之前
221             for (int i = 0; i <= 2; i++) { if (!A_Check->t[i]) t0++;  }
222             if (t0 != 1 ) {//细节 判断是否有大人
223                 if (!A_Check->t[i])
224                     A_Check->t[i] = 1;
225                 else continue;
226                 for (int i = 0; i <= 8; i++)  t1[i] = A_Check->t[i];//操作之后
227                 for (int j = 0; j <= 7; j++) {
228                     for (int i = 0; i <= 8; i++)  A_Check->t[i] = t1[i];//操作之前
229                     if (i == j) continue;
230                     if (!A_Check->t[j])
231                         A_Check->t[j] = 1;
232                     else continue;
233                     //Astar A_Check_Check;//判断对面的是否也符合 
234                     if (Is_OK(A_Check)) {
235                         Astar Sstar = (Astar)malloc(sizeof(AS));
236                         for (int i = 0; i <= 8; i++) Sstar->t[i] = A_Check->t[i];
237                         Sstar->t[8] = 1;//细节必须在IF的上面
238                         if (Is_SomeBefor(Sstar, A_Check->parents) || Is_Same(A_Check_L, Sstar)) free(Sstar);
239                         else
240                         {
241                             Sstar->chilren = NULL;
242                             Sstar->parents = A_Check; 
243                             Sstar->next = NULL;
244                             Sstar->f = G_value(Sstar, A_Check) + H_value(Sstar);
245                             for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];// cout << Sstar->f<<"++"<<Sstar->parents->g<<"++"<<Sstar->h<<" ";
246                             Enter_Close(Sstar, A_Check_L); //for (int i = 0; i <= 8; i++) cout << A_Check->t[i] << "<<++<"; cout << endl;
247                         }
248                     }
249                 }
250             }
251             else break;
252         }
253         for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];
254         for (int i = 0; i <= 2; i++) {
255           for (int i = 0; i <= 8; i++)  A_Check->t[i] = t2[i];
256             if (!A_Check->t[i])
257                 A_Check->t[i] = 1;
258             else continue;
259             if (Is_OK(A_Check))
260             {
261                 Astar Sstar1 = (Astar)malloc(sizeof(AS));
262                 for (int i = 0; i <= 8; i++) Sstar1->t[i] = A_Check->t[i];            
263                 Sstar1->t[8] = 1;
264                 if (Is_SomeBefor(Sstar1, A_Check->parents) || Is_Same(A_Check_L, Sstar1)) free(Sstar1);
265                 else 
266                 {
267                     Sstar1->chilren = NULL;
268                     Sstar1->parents = A_Check;
269                     Sstar1->next = NULL;
270                     Sstar1->f = G_value(Sstar1, A_Check) + H_value(Sstar1);
271                     for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];// cout << Sstar1->f << "--" << Sstar1->g << "--" << Sstar1->h << " ";
272                     Enter_Close(Sstar1, A_Check_L);
273                 }
274             }
275         }
276     }
277     
278     else {
279         for (int i = 0; i <= 2; i++) {
280             for (int i = 0; i <= 8; i++) A_Check->t[i]=t2[i];
281             {
282                 if (A_Check->t[i])
283                     A_Check->t[i] = 0;
284                 else continue; 
285             }
286             
287             if (Is_OK(A_Check)) {
288                 Astar Sstar = (Astar)malloc(sizeof(AS));
289                 for (int i = 0; i <= 8; i++) Sstar->t[i] = A_Check->t[i];
290                 Sstar->t[8] = 0;
291                 if (Is_SomeBefor(Sstar, A_Check->parents) || Is_Same(A_Check_L, Sstar)) free(Sstar);
292                 else 
293                 {
294                     Sstar->chilren = NULL;
295                     Sstar->parents = A_Check;
296                     Sstar->next = NULL;
297                     Sstar->f = G_value(Sstar, A_Check) + H_value(Sstar);
298                     for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i]; //cout << Sstar->f << "**" << Sstar->g << "**" << Sstar->h << " ";
299                     Enter_Close(Sstar, A_Check_L);// for (int i = 0; i <= 8; i++)   //cout<< Sstar->t[i]<<"<<<"; cout << endl;
300                 }
301             }
302             //for (int i = 0; i <= 8; i++)  A_Check->t[i]=t2[i];
303         }
304         for (int i = 0; i <= 8; i++) A_Check->t[i]=t2[i] ;
305         A_Check->t[8] = 1;
306         for (int i = 0; i <= 2; i++) {
307           for (int i = 0; i <= 8; i++)  A_Check->t[i]=t2[i];
308             if (!A_Check->t[i]) continue;
309             else
310                 A_Check->t[i] = 0; 
311             for (int i = 0; i <= 8; i++)  t1[i] = A_Check->t[i];
312             for (int j = 0; j <= 3; j++) {
313                 if (i==j) continue;
314                 for (int i = 0; i <= 8; i++)  A_Check->t[i]=t1[i];
315                 if (!A_Check->t[j]) continue;
316                 else
317                     A_Check->t[j] = 0;
318                 if (Is_OK(A_Check)) {
319                     Astar Sstar1 = (Astar)malloc(sizeof(AS));
320                     for (int i = 0; i <= 8; i++) Sstar1->t[i] = A_Check->t[i]; 
321                     Sstar1->t[8] = 0; 
322                     if (Is_SomeBefor(Sstar1, A_Check->parents )|| Is_Same(A_Check_L, Sstar1)) free(Sstar1);
323                     else 
324                     {
325                         Sstar1->chilren = NULL;
326                         Sstar1->parents = A_Check;
327                         Sstar1->next = NULL;
328                         Sstar1->f = G_value(Sstar1, A_Check) + H_value(Sstar1); 
329                         for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];// cout << Sstar1->f << "//" << Sstar1->g << "//" << Sstar1->h << " ";
330                         Enter_Close(Sstar1, A_Check_L);// for (int i = 0; i <= 8; i++)  // cout<< Sstar1->t[i]<<"<<<"; cout << endl;
331                     }
332                 }
333             }
334         }
335     }for (int i = 0; i <= 8; i++) A_Check->t[i] = t2[i];
336 }
337 int Is_Over(Astar Over) {
338     int i;
339     for (i = 0; i <= 8; i++)
340         if (!Over->t[i]) {
341             break;
342         }
343     if (i == 9)   return 1; 
344     return 0;
345 }
346 void A_Star(int begin[],int end[]) {//核心 1
347     Astar A_Check=NULL;
348     Astar Check_A=NULL;
349     Astar A_Check_L;
350     Astar A_Check_L_One;
351     //Astar A_ChuanChuan;//串串儿 一串到底
352     Astar M = NULL;
353     Astar N = NULL;
354     //A_ChuanChuan = NULL;
355     int F=0,f=0;
356     create(Open);
357     create(Close);
358     create(A_Check_L);
359     Open_EnterFirst(begin,end); 
360     while (Is_Empety(Open)) {
361         Out(Open, A_Check); 
362         Enter_Close(A_Check, Close); 
363         if (Is_Over(A_Check)) { F = 1; break; }
364         A_Check_LS(A_Check_L, A_Check);
365         /*    
366         while (Is_Empety(A_Check_L))
367         {
368             Out(A_Check_L, A_Check_L_One);cout<< "++"; //测验
369         }
370         */
371         while (Is_Empety(A_Check_L))
372         {
373             Out(A_Check_L, A_Check_L_One);
374             if (Is_In(A_Check_L_One, Open, M, N))
375             {
376                 
377                 if (M->g > A_Check_L_One->g)
378                 {
379                     M->parents = A_Check_L_One->parents;
380                     M->g = A_Check_L_One->g;
381                     M->f = A_Check_L_One->f;
382                 }
383             }
384             else if (Is_In(A_Check_L_One, Close, M, N))
385             {
386                 if (M->g > A_Check_L_One->g)
387                 {
388                     
389                     Astar Q; 
390                     M->parents = A_Check_L_One->parents;
391                     M->g = A_Check_L_One->g;
392                     M->f = A_Check_L_One->f;    
393                     M->chilren = NULL;
394                     Out(Close, Q);
395                     Add_Open(Open, Q);
396                 }
397             }
398             else 
399             {
400                 
401                 Add_Open(Open, A_Check_L_One); 
402             }
403         }
404         
405     } if(F)
406             Out_Path(A_Check); 
407 }
408 typedef struct A {
409     int t[9];
410     struct A* next;
411     struct A* parents;
412     struct AA* chilren;
413     int f;
414     int g;
415     int h;
416 } AS, * Astar;
417 typedef struct AA {
418     struct AA* next;
419     struct A* To_A;;
420 }AAS,*AAstar;
421 Astar Open;
422 Astar Close;






 








第1步：猎人和狗过河
第2步：猎人回来
第3步：猎人带一个女儿过河
第4步：猎人和狗回来
第5步：女人带另外一个女孩过河
第6步：女人回来
第7步：男人和女人过河(还要把船带回来，所以男人得过去)
第8步：男人过来
第9步：猎人和狗过去
第10步：女人回来
第11步：女人和男人过河
第12步：男人回来
第13步：男人带一个男孩过河
第14步：猎人和狗回去
第15步：猎人和男孩过河
第16步：猎人回去
第17步：猎人和狗过来


这个猎人渡河问题本身挺简单的，但要用A*算法实现起来并不太简单，因为细节实在太多。另外这题的转态结点在OPEN表和CLOUED表中的并不多，这使得一些Astar算法中的必须要的操作在这里并没有起到作用，所以这个并不是Astar算法的一个典型问题。那为啥我还要写这个呢？我们小唐老师在讲Astar算法时说谁要是用Astar算法写出这个，谁期末就不用考了。所以我就搞起来了，，。




 







		

		
posted @ 
2020-04-13 22:24 
VL_L_^_^ 
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