BFS解决八数码问题和狼人过河问题

Posted 2021-01-13 eastblue

1、八数码问题

问题描述：

初态：

0    1    2

3    4    5

6    7    8

如何移动交换0的位置达到终态

1    2     3

4    5     6

7    8     0

 

思路如下：

先将图转换为一个整数

初态：
876543210
终态：
087654321

构造状态的数据结构

struct node{
int x;
int where0;
}

运动规则如下

switch where0:
case0: d,r
case1: d,l,r
case2: d,l
case3: u,d,r
case4: u,d,l,r
case5: u,d,l
case6: u,r
case7: u,l,r
case8: u,l

 

switch dir:
case u:t=x/10^(where0-3)%10; x=x-10^(where0-3)*t+10^where0*t;
case d:t=x/10^(where0+3)%10; x=x-10^(where0+3)*t+10^where0*t;
case l:t=x/10^(where0-1)%10; x=x-10^(where0-1)*t+10^where0*t;
case r:t=x/10^(where0+1)%10; x=x-10^(where0+1)*t+10^where0*t;

代码：

  1 #include<iostream>
  2 #include<map>
  3 #include<queue>
  4 #include<cstdlib> 
  5 using namespace std;
  6 
  7 int pow10[10]={1,10,100,1000,10000,100000,
  8 1000000,10000000,100000000,1000000000};
  9 
 10 //数据结构 
 11 struct node{
 12     int x;//表示当前状态图 
 13     int where0;//0的位置 
 14     struct node *pre;//父节点 
 15 };
 16 
 17 //运动规则 
 18 node * goAction(node *p,int dir){
 19     int x=p->x,where0=p->where0;
 20     node *ans=(node *)malloc(sizeof(node));
 21     ans->pre=p;
 22     int t;
 23     switch(dir){
 24         case 1://up
 25             t=x/pow10[where0-3]%10;
 26             x=x-pow10[where0-3]*t+pow10[where0]*t;
 27             where0-=3;
 28             break;
 29         case 2://down
 30             t=x/pow10[where0+3]%10;   
 31             x=x-pow10[where0+3]*t+pow10[where0]*t;
 32             where0+=3;
 33             break;
 34         case 3://left
 35             t=x/pow10[where0-1]%10;   
 36             x=x-pow10[where0-1]*t+pow10[where0]*t;
 37             where0-=1;
 38             break;
 39         case 4://right
 40             t=x/pow10[where0+1]%10;   
 41             x=x-pow10[where0+1]*t+pow10[where0]*t;
 42             where0+=1;
 43             break;
 44     }
 45     ans->x=x;
 46     ans->where0=where0;
 47     return ans;    
 48 }
 49 
 50 queue<node *>nq;//状态队列 
 51 map<int,int>nm;//判重 
 52 
 53 //新节点加入队列 
 54 int join(node *a){
 55     if(nm[a->x]==1)//重复节点不加入队列 
 56         return 0;
 57     if(a->x==87654321){//抵达终态 
 58         cout<<"路径："<<endl; 
 59         node *h=a;
 60         int step=0;
 61         while(h!=NULL)//打印路径和步数 
 62         {
 63             cout<<h->x<<"  ";
 64             step++;
 65             h=h->pre;        
 66         }
 67         cout<<step<<endl;
 68         return 1;
 69     }
 70     nm[a->x]=1;
 71     nq.push(a);//加入队列 
 72     return 0;
 73 }
 74 
 75 void fun(){
 76     
 77     while(!nq.empty()){
 78         node *p=nq.front();
 79         nq.pop();
 80         switch(p->where0){//运动规则 
 81             case 0:
 82                 join(goAction(p,2));
 83                 join(goAction(p,4));
 84                 break;
 85             case 1:
 86                 join(goAction(p,2));
 87                 join(goAction(p,3));    
 88                 join(goAction(p,4));
 89                 break;
 90             case 2:
 91                 join(goAction(p,2));
 92                 join(goAction(p,3));
 93                 break;
 94             case 3:
 95                 join(goAction(p,1));
 96                 join(goAction(p,2));
 97                 join(goAction(p,4));
 98                 break;
 99             case 4:
100                 join(goAction(p,1));
101                 join(goAction(p,2));
102                 join(goAction(p,3));
103                 join(goAction(p,4));
104                 break;
105             case 5:
106                 join(goAction(p,1));
107                 join(goAction(p,2));
108                 join(goAction(p,3));
109                 break;
110             case 6:
111                 join(goAction(p,1));
112                 join(goAction(p,4));
113                 break;
114             case 7:
115                 join(goAction(p,1));
116                 join(goAction(p,3));
117                 join(goAction(p,4));
118                 break;
119             case 8:
120                 join(goAction(p,1));
121                 join(goAction(p,3));
122                 break;
123 
124         }
125             
126     }
127      
128 }
129 
130 int main()
131 {
132     node *begin=(node *)malloc(sizeof(node));//初始状态 
133     begin->x=876543210;
134     begin->where0=0;
135     begin->pre=NULL;
136     join(begin);
137     fun();
138 
139 
140 }

2、狼人过河问题

问题描述：

{wolf,human,boat}
分别代表右岸的狼，人，船的数目

初态：{3，3，1}
终态：{0，0，0}

如何从初态抵达终态

规则：
1、每次过河船上可以一人或两人
switch boat:
case 0:
boat++;
wolf+=1||wolf+=2||human+=1||human+=2||wolf+=1,human+=1;
case 1:
boat--;
wolf-=1||wolf-=2||human-=1||human-=2||wolf-=1,human-=1;
2、两岸的狼不能比人多（人数不为0时）
no1：wolf<0||wolf>3||human<0||human>3||boat<0||boat>1
no2:(human>0&&human<wolf)||((3-human)>0&&(3-human)<(3-wolf))

 

数据结构

struct node{
int wolf;
int human;
int boat;
struct node *pre;
};

 

代码：

  1 #include<iostream>
  2 #include<map>
  3 #include<queue>
  4 #include<cstdlib> 
  5 #include<String>
  6 using namespace std;
  7 
  8 struct node{
  9 int wolf;
 10 int human;
 11 int boat;
 12 struct node *pre;
 13 void init(int a,int b,int c,struct node *p){
 14     this->wolf=a;
 15     this->human=b;
 16     this->boat=c;
 17     this->pre=p;
 18 }
 19 
 20 bool operator < (const node x) const{//重载运算符，注意map是基于红黑树实现，每个节点需要具备可比性 
 21     int hash1=this->wolf*100+this->human*10+this->boat;
 22     int hash2=x.wolf*100+x.human*10+x.boat;
 23     return hash1<hash2;
 24 }
 25 
 26 
 27 
 28 };
 29 
 30 queue<node *>nq;//状态队列 
 31 map<node,int>nm;//状态判重 
 32 
 33 //判断能否加入队列 
 34 int join(node *a){
 35     if(nm[*a]==1)
 36         return 0;
 37     int wolf=a->wolf,human=a->human,boat=a->boat;
 38     if(wolf<0||wolf>3||human<0||human>3||boat<0||boat>1)
 39         return 0;
 40     if((human>0&&human<wolf)||((3-human)>0&&(3-human)<(3-wolf)))
 41         return 0;
 42     if(a->wolf==0&&a->human==0&&a->boat==0){//终态 
 43         node *h=a;
 44         int step=0;
 45         while(h!=NULL){//打印路径和步数 
 46             step++;
 47             cout<<"{ "<<h->wolf<<" , "<<h->human<<" , "<<h->boat<<" }  ";
 48             h=h->pre;
 49         }
 50         cout<<endl;
 51         cout<<step<<endl;
 52         return 1;
 53     }
 54     nm[*a]=1;
 55     nq.push(a);
 56     return 0;
 57 }
 58 
 59 //运动规则 
 60 void goAction(node *p){
 61     node *a=(node *)malloc(sizeof(node));
 62     node *b=(node *)malloc(sizeof(node));
 63     node *c=(node *)malloc(sizeof(node));
 64     node *d=(node *)malloc(sizeof(node));
 65     node *e=(node *)malloc(sizeof(node));
 66     switch(p->boat){
 67         case 0://注意，不能在case里新建变量 
 68             
 69             a->init(p->wolf+1,p->human,1,p);
 70             join(a);
 71             
 72             b->init(p->wolf+2,p->human,1,p);
 73             join(b);
 74             
 75             c->init(p->wolf,p->human+1,1,p);
 76             join(c);
 77             
 78             d->init(p->wolf,p->human+2,1,p);
 79             join(d);
 80             
 81             e->init(p->wolf+1,p->human+1,1,p);
 82             join(e);
 83             break;
 84         case 1:
 85             
 86             a->init(p->wolf-1,p->human,0,p);
 87             join(a);
 88             
 89             b->init(p->wolf-2,p->human,0,p);
 90             join(b);
 91             
 92             c->init(p->wolf,p->human-1,0,p);
 93             join(c);
 94             
 95             d->init(p->wolf,p->human-2,0,p);
 96             join(d);
 97             
 98             e->init(p->wolf-1,p->human+1,0,p);
 99             join(e);
100             break;
101     }
102     return;
103 }
104 
105 
106 
107 void fun(){
108     while(!nq.empty()){
109         node *p=nq.front();
110         nq.pop();
111         goAction(p);
112     }
113     
114 }
115 
116 int main(){
117     node *begin=(node *)malloc(sizeof(node));//初态 
118     begin->init(3,3,1,NULL);
119     join(begin);
120     fun();
121 
122 }