Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10778 Accepted Submission(s):
2873
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
Source
Recommend
康托展开优化，代码自己写的过了，做EIGHTII的时候，发现别人bfs（）很简单，而且map[4][2]写的看不懂。
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<cstdlib> 5 #include<string> 6 #include<queue> 7 using namespace std; 8 9 struct node 10 { 11 bool flag; 12 char str; 13 int father; 14 }hash[363001]; 15 struct st 16 { 17 char t[10]; 18 }; 19 queue<st>Q; 20 int ans[10]={1}; 21 22 int ktzk(char *c) 23 { 24 int i,j,k,sum=0; 25 for(i=0;i<=8;i++) 26 { 27 k=0; 28 for(j=i+1;j<=8;j++) 29 if(c[i]>c[j]) 30 k++; 31 sum=sum+k*ans[8-i]; 32 } 33 return sum; 34 } 35 void bfs() 36 { 37 int i,k,num,val; 38 struct st cur,t; 39 k=ktzk("123456780"); 40 hash[k].flag=true; 41 hash[k].str=' '; 42 hash[k].father=k; 43 44 strcpy(t.t,"123456780"); 45 Q.push(t); 46 47 while(!Q.empty()) 48 { 49 cur=Q.front(); 50 Q.pop(); 51 val=ktzk(cur.t); 52 for(i=0;i<=8;i++) 53 { 54 if(cur.t[i]=='0') 55 { 56 k=i; 57 break; 58 } 59 } 60 if(k!=2 && k!=5 && k!=8)//rigth 61 { 62 t=cur; 63 swap(t.t[k],t.t[k+1]); 64 num=ktzk(t.t); 65 if(hash[num].flag==false) 66 { 67 hash[num].flag=true; 68 hash[num].str='r'; 69 hash[num].father=val; 70 Q.push(t); 71 } 72 } 73 if(k!=0 && k!=3 && k!=6)//left 74 { 75 t=cur; 76 swap(t.t[k],t.t[k-1]); 77 num=ktzk(t.t); 78 if(hash[num].flag==false) 79 { 80 hash[num].flag=true; 81 hash[num].str='l'; 82 hash[num].father=val; 83 Q.push(t); 84 } 85 } 86 if(k>=3)//u 87 { 88 t=cur; 89 swap(t.t[k],t.t[k-3]); 90 num=ktzk(t.t); 91 if(hash[num].flag==false) 92 { 93 hash[num].flag=true; 94 hash[num].str='u'; 95 hash[num].father=val; 96 Q.push(t); 97 } 98 } 99 if(k<=5)//D 100 { 101 t=cur; 102 swap(t.t[k],t.t[k+3]); 103 num=ktzk(t.t); 104 if(hash[num].flag==false) 105 { 106 hash[num].flag=true; 107 hash[num].str='d'; 108 hash[num].father=val; 109 Q.push(t); 110 } 111 } 112 } 113 } 114 void prepare() 115 { 116 int i; 117 for(i=0;i<=363000;i++) 118 { 119 hash[i].flag=false; 120 } 121 for(i=1;i<=9;i++) ans[i]=ans[i-1]*i; 122 bfs(); 123 } 124 int main() 125 { 126 prepare(); 127 char a[50],b[10],c[100]; 128 bool tom[10]; 129 int i,j,k,cur; 130 while(gets(a)) 131 { 132 k=strlen(a); 133 b[9]=' '; 134 for(i=0,j=0;i<k;i++) 135 { 136 if(a[i]=='x' || (a[i]>='1'&&a[i]<='8')) 137 { 138 if(a[i]=='x') 139 b[j]='0'; 140 else b[j]=a[i]; 141 j++; 142 } 143 } 144 memset(tom,false,sizeof(tom)); 145 for(i=0;i<=8;i++) 146 { 147 tom[b[i]-'0']=true; 148 } 149 for(i=0;i<=8;i++) 150 { 151 if(tom[i]==false) 152 break; 153 } 154 if(i<=8){printf("unsolvable ");continue;} 155 cur=ktzk(b); 156 if(hash[cur].flag==false) 157 { 158 printf("unsolvable "); 159 continue; 160 } 161 k=0; 162 while(hash[cur].father!=cur) 163 { 164 c[k]=hash[cur].str; 165 k++; 166 cur=hash[cur].father; 167 } 168 for(i=0;i<=k-1;i++) 169 { 170 if(c[i]=='u')printf("d"); 171 if(c[i]=='d')printf("u"); 172 if(c[i]=='l')printf("r"); 173 if(c[i]=='r')printf("l"); 174 } 175 printf(" "); 176 } 177 return 0; 178 }