HDU1043 Eight(BFS)
Eight(South Central USA 1998)
Time Limit:5000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
Input
You will receive, several descriptions of configuration of the 8 puzzle. One description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
Output
You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line. Do not print a blank line between cases.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
题目简单翻译：
八数码
给你一个八数码：（格式是一个九宫格，x代表空），问怎么操作能达到目标，即：
1 2 3
4 5 6
7 8 x
u代表空格向上交换，d代表空格向下交换，l代表空格向左交换，r代表空格向右交换。
例如：
给你一个八数码：
1 2 3
x 4 6
7 5 8
则把它变成目标需要三步：
1 2 3 r 1 2 3 d 1 2 3 r 1 2 3
x 4 6 ---> 4 x 6 ---> 4 5 6 ---> 4 5 6
7 5 8 7 5 8 7 x 8 7 8 x
所以这个样例应该输出：
rdr
如果不能到达目标就输出“unsolvable”.
解题思路：广度优先搜索（BFS）
因为是多组数据，我们可以先求出所有情况，然后每次询问的时候直接输出结果就好了。
求出所有结果，我们就可以根据结果，来逆向bfs，直到所有的情况都求到。
代码：
1 #include<cstdio> 2 #include<string> 3 #include<queue> 4 #include<cstring> 5 using namespace std; 6 struct node 7 { 8 int t[3][3],x,y,Can; 9 int Last_Can,dir; 10 } St[370000]; 11 int fac[]= {1,1,2,6,24,120,720,5040,40320}; 12 //康托展开的数组 13 //康托展开就是把一组数据按照字典序排列的那组数据的序号 14 15 int vis[370000]; 16 queue<int> Q; 17 char dr[]="rlud"; 18 int dx[]= {0,0,1,-1}; 19 int dy[]= {-1,1,0,0}; 20 //方向数组，与实际的方向相反，因为是逆向操作 21 int Cantor(int *t)//对一组数据求康拓值 22 { 23 int rev=0; 24 for(int i=0; i<9; i++) 25 { 26 int counted=0; 27 for(int j=i+1; j<9; j++) 28 if(t[i]>t[j]) counted++; 29 rev+=counted*fac[8-i]; 30 } 31 return rev; 32 } 33 bool check(int x,int y) //检查这个点是不是在矩形内 34 { 35 return x>=0&&x<3&&y>=0&&y<3; 36 } 37 void solve()//bfs求出所有的情况，并储存下来父节点 38 { 39 while(!Q.empty()) Q.pop(); 40 node st; 41 st.x=st.y=2; 42 int s[3][3]= {1,2,3,4,5,6,7,8,0}; 43 int t[9]; 44 for(int i=0; i<3; i++) 45 for(int j=0; j<3; j++) 46 t[i*3+j]=s[i][j]; 47 for(int i=0; i<3; i++) 48 for(int j=0; j<3; j++) 49 st.t[i][j]=s[i][j]; 50 int StCan=Cantor(t); 51 st.Can=StCan; 52 st.Last_Can=-1; 53 st.dir=-1; 54 memset(vis,0,sizeof vis); 55 vis[StCan]=1; 56 St[StCan]=st; 57 Q.push(StCan); 58 int Sum=0; 59 while(!Q.empty()) 60 { 61 Sum++; 62 int TempCan=Q.front(); 63 Q.pop(); 64 for(int i=0; i<4; i++) 65 { 66 node e=St[TempCan]; 67 int curx=e.x+dx[i]; 68 int cury=e.y+dy[i]; 69 if(check(curx,cury)) 70 { 71 int c=e.t[curx][cury]; 72 e.t[curx][cury]=e.t[e.x][e.y]; 73 e.t[e.x][e.y]=c; 74 e.x=curx; 75 e.y=cury; 76 int t[9]; 77 for(int i=0; i<3; i++) 78 for(int j=0; j<3; j++) 79 t[i*3+j]=e.t[i][j]; 80 e.Can=Cantor(t); 81 e.Last_Can=TempCan; 82 e.dir=i; 83 if(!vis[e.Can]) 84 { 85 vis[e.Can]=1; 86 St[e.Can]=e; 87 Q.push(e.Can); 88 } 89 } 90 } 91 } 92 } 93 int main() 94 { 95 char c[10]; 96 int t[9],x=0; 97 solve(); 98 while(scanf("%s",c)!=EOF) 99 { 100 if(c[0]=='x') c[0]='0'; 101 t[0]=c[0]-'0'; 102 for(int i=1;i<9;i++) 103 { 104 scanf("%s",c); 105 if(c[0]=='x') c[0]='0'; 106 t[i]=c[0]-'0'; 107 } 108 int AnsCan=Cantor(t); 109 if(vis[AnsCan]) 110 { 111 int p=AnsCan; 112 while(St[p].Last_Can+1) 113 { 114 printf("%c",dr[St[p].dir]); 115 p=St[p].Last_Can; 116 } 117 printf("\n"); 118 } 119 else 120 printf("unsolvable\n"); 121 } 122 return 0; 123 }