hdu_1043 eight

 1 #include<iostream>
 2 #include<string>
 3 #include<queue>
 4 #include<string.h>
 5 using namespace std;
 6 
 7 const int maxn=1000000;
 8 
 9 bool vis[maxn];      //是否被访问过
10 string path[maxn];   //记录路径
11 int fac[]={          //阶乘
12     1,1,2,6,24,120,720,5040,40320,362880
13 };
14 
15 int contor(int s[]){ //康托展开
16     int sum=0;
17     for(int i=0;i<9;i++){
18         int num=0;
19         for(int j=i+1;j<9;j++)
20             if(s[j] < s[i] )
21                 num ++;
22         sum += (num * fac[8-i]);
23     }
24 
25     return sum;
26 }
27 struct node{        //节点
28     int s[9];       //记录状态
29     int cur;        //记录当前位置
30     string path;    //记录路径
31 };
32 
33 queue<node> q;
34 int move[4][2]={    //移动方向
35     -1,0,1,0,0,-1,0,1
36 };
37 
38 char mop[5]="durl"; // 这里的和上面的移动方向相反，是为了反向搜索
39 
40 void bfs(){
41     node no,next;
42     for(int i=0;i<8;i++) no.s[i]=i+1; //将起点初始化为目标，进行反向bfs
43     no.s[8]=0;no.cur=8;                  
44     no.path="";
45     q.push(no);
46     vis[contor(no.s)]=true;
47     path[contor(no.s)]="";
48     while(!q.empty()){
49         no=q.front();q.pop();
50         int x=no.cur/3;
51         int y=no.cur%3;
52         for(int i=0;i<4;i++){
53             int tx=x+move[i][0];
54             int ty=y+move[i][1];
55             if(tx<0 || tx>2 || ty <0 || ty >2) continue;
56             next=no;next.cur=3*tx+ty;
57             next.s[no.cur] = no.s[next.cur];
58             next.s[next.cur] =0;
59             int val=contor(next.s);
60             if(!vis[val]){
61                 vis[val] =true;
62                 next.path=mop[i] + next.path;
63                 q.push(next);
64                 path[val] = next.path;
65             }
66         }
67     }
68 }
69 
70 
71 int main(){
72     char ch;
73     node no;
74     bfs();
75     while(cin >> ch){
76         if(ch == ‘x‘) {no.s[0] = 0; no.cur=0;}
77         else          no.s[0] =ch - ‘0‘;
78         for(int i=1;i<9;i++){
79             cin>>ch;
80             if(ch == ‘x‘){
81                 no.s[i] = 0; no.cur=i;
82             }else 
83                 no.s[i] = ch - ‘0‘;
84         }
85         int val=contor(no.s);
86         if(vis[val]){
87             cout<<path[val]<<endl;
88         }
89         else 
90             cout<<"unsolvable"<<endl;
91     }
92     return 0;
93 }