Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11226 Accepted Submission(s): 3013
Special Judge
Problem 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
1 #include <iostream> 2 #include <string.h> 3 #include <stdio.h> 4 #include <math.h> 5 #include <stdlib.h> 6 #include <queue> 7 using namespace std; 8 typedef struct eight 9 { 10 char a[3][3]; 11 int statu,nowx,nowy; 12 } eight; 13 eight s,e; 14 int f[12],flag[500000],ok; 15 int father1[500000],father2[500000]; 16 int move1[500000],move2[500000],last; 17 queue<eight>q1,q2; 18 void init() 19 { 20 f[0]=1; 21 int i; 22 for(i=1; i<12; i++)f[i]=f[i-1]*i; 23 } 24 void work(eight &a) 25 { 26 int i,j,k=0,ans=0,t; 27 char b[9]; 28 for(i=0; i<3; i++) 29 { 30 for(j=0; j<3; j++) 31 { 32 if(a.a[i][j]=='x')a.nowx=i,a.nowy=j; 33 b[k++]=a.a[i][j]; 34 } 35 } 36 k=8; 37 for(i=0; i<9; i++) 38 { 39 t=0; 40 for(j=i+1; j<9; j++) 41 if(b[j]<b[i])t++; 42 ans+=f[k--]*t; 43 } 44 a.statu=ans; 45 } 46 void copy(eight &a,eight &b) 47 { 48 int i,j; 49 for(i=0; i<3; i++) 50 for(j=0; j<3; j++)a.a[i][j]=b.a[i][j]; 51 } 52 int w[4][2]= {{0,1},{0,-1},{1,0},{-1,0}}; 53 void ex_BFS(eight &now,int n,int check) 54 { 55 int i,r,c; 56 eight in; 57 for(i=0; i<4; i++) 58 { 59 60 r=now.nowx+w[i][0]; 61 c=now.nowy+w[i][1]; 62 if(r>=0&&r<3&&c>=0&&c<3) 63 { 64 copy(in,now); 65 swap(in.a[r][c],in.a[now.nowx][now.nowy]); 66 work(in); 67 if(flag[in.statu]!=n) 68 { 69 if(n==1) 70 { 71 q1.push(in); 72 father1[in.statu]=now.statu; 73 move1[in.statu]=i; 74 } 75 else 76 { 77 q2.push(in); 78 father2[in.statu]=now.statu; 79 move2[in.statu]=i; 80 } 81 if(flag[in.statu]==check) 82 { 83 ok=1; 84 last=in.statu; 85 return ; 86 } 87 flag[in.statu]=n; 88 } 89 } 90 } 91 } 92 bool TBFS(eight &s,eight &e) 93 { 94 memset(flag,0,sizeof(flag)); 95 memset(move1,-1,sizeof(move1)); 96 memset(move2,-1,sizeof(move2)); 97 memset(father1,-1,sizeof(father1)); 98 memset(father2,-1,sizeof(father2)); 99 eight now; 100 ok=0; 101 while(!q1.empty())q1.pop(); 102 while(!q2.empty())q2.pop(); 103 q1.push(s),flag[s.statu]=1; 104 q2.push(e),flag[e.statu]=2; 105 while((!q1.empty())||(!q2.empty())) 106 { 107 now=q1.front(); 108 q1.pop(); 109 ex_BFS(now,1,2); 110 if(ok)return 1; 111 112 now=q2.front(); 113 q2.pop(); 114 ex_BFS(now,2,1); 115 if(ok)return 1; 116 } 117 return 0; 118 } 119 bool check1(eight &s) 120 { 121 int i,j,k=0; 122 char b[9]; 123 for(i=0; i<3; i++) 124 for(j=0; j<3; j++) 125 b[k++]=s.a[i][j]; 126 int ans=0; 127 for(i=0; i<9; i++) 128 { 129 if(b[i]!='x') 130 for(j=0; j<i; j++) 131 if(b[j]!='x'&&b[i]<b[j])ans++; 132 } 133 return (ans&1); 134 } 135 void printpath() 136 { 137 deque<char>q; 138 while(!q.empty())q.pop_back(); 139 int now=last; 140 while(father1[now]!=-1) 141 { 142 if(move1[now]==0)q.push_back('r'); 143 else if(move1[now]==1)q.push_back('l'); 144 else if(move1[now]==2)q.push_back('d'); 145 else if(move1[now]==3)q.push_back('u'); 146 now=father1[now]; 147 } 148 while(!q.empty()) 149 { 150 cout<<q.back(); 151 q.pop_back(); 152 } 153 154 now=last; 155 while(father2[now]!=-1) 156 { 157 if(move2[now]==0)cout<<'l'; 158 else if(move2[now]==1)cout<<'r'; 159 else if(move2[now]==2)cout<<'u'; 160 else if(move2[now]==3)cout<<'d'; 161 now=father2[now]; 162 } 163 } 164 int main() 165 { 166 int i,j; 167 init(); 168 char w; 169 while(cin>>w) 170 { 171 char an='1'; 172 for(i=0; i<3; i++) 173 for(j=0; j<3; j++) 174 { 175 if(i||j) 176 cin>>s.a[i][j]; 177 e.a[i][j]=an++; 178 } 179 s.a[0][0]=w; 180 181 e.a[2][2]='x'; 182 if(check1(s)) 183 { 184 cout<<"unsolvable"<<endl; 185 continue; 186 } 187 work(s); 188 work(e); 189 if(s.statu==e.statu) 190 { 191 cout<<endl; 192 continue; 193 } 194 if(TBFS(s,e)) 195 { 196 printpath(); 197 cout<<endl; 198 } 199 else 200 cout<<"unsolvable"<<endl; 201 } 202 203 }