Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25454 Accepted Submission(s):
6781
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
bfs+优先队列
1 # include<string.h> 2 # include<queue> 3 # define N 363000 4 using namespace std; 5 bool visit[N]; 6 char visit1[N]; 7 int pre[N],st,a[10],ed; 8 int dir[9]={1,1,2,6,24,120,720,5040,40320}; 9 struct node{ 10 int ma[10]; 11 int ans1; 12 int x; 13 int f; 14 int g; 15 bool operator <(const node &a)const { 16 return a.f < f;//优先访问f较小者 17 } 18 }; 19 int hsh(int s[]) 20 { 21 int i,j,cnt,sum; 22 sum=0; 23 for(i=1;i<=9;i++) 24 { 25 cnt=0; 26 for(j=1;j<i;j++) 27 if(s[j]>s[i]) cnt++; 28 sum+=cnt*dir[i-1]; 29 } 30 return sum; 31 } 32 int ABS(int x) {return x<0?(-x):x;} 33 int h(int s[])//不算x时的曼哈顿距离 34 { 35 int curx,cury,endx,endy,sum,i,ans; 36 sum=0; 37 for(i=1;i<=9;i++) 38 { 39 if(s[i]==9) continue; 40 ans=s[i]; 41 curx=(i+2)/3; 42 cury=(i-1)%3+1; 43 endx=(ans+2)/3; 44 endy=(ans-1)%3+1; 45 sum=sum+ABS(curx-endx)+ABS(cury-endy); 46 } 47 return sum; 48 } 49 void bfs() 50 { 51 int ans,i; 52 priority_queue<node>q; 53 node cur,next; 54 cur.ans1=st=hsh(a); 55 visit[cur.ans1]=1; 56 if(st==ed) return; 57 for(i=1;i<=9;i++) 58 { 59 cur.ma[i]=a[i]; 60 if(a[i]==9) cur.x=i; 61 } 62 cur.g=0;//表示深度 63 cur.f=h(a); 64 q.push(cur); 65 while(!q.empty()) 66 { 67 cur=q.top(); 68 q.pop(); 69 if((cur.x+2)/3!=1) //向上翻 70 { 71 next=cur; 72 next.x=cur.x-3; 73 next.ma[cur.x]=next.ma[next.x]; 74 next.ma[next.x]=9; 75 ans=hsh(next.ma); 76 if(!visit[ans]) 77 { 78 next.g++; 79 next.f=next.g+h(next.ma); 80 visit[ans]=1; 81 next.ans1=ans; 82 pre[ans]=cur.ans1; 83 visit1[ans]='u'; 84 if(ans==ed) return; 85 q.push(next); 86 } 87 } 88 if((cur.x+2)/3!=3)//向下翻 89 { 90 next=cur; 91 next.x=cur.x+3; 92 next.ma[cur.x]=next.ma[next.x]; 93 next.ma[next.x]=9; 94 ans=hsh(next.ma); 95 if(!visit[ans]) 96 { 97 next.g++; 98 next.f=next.g+h(next.ma); 99 visit[ans]=1; 100 next.ans1=ans; 101 pre[ans]=cur.ans1; 102 visit1[ans]='d'; 103 if(ans==ed) return; 104 q.push(next); 105 } 106 } 107 if(cur.x%3!=1)//向左翻 108 { 109 next=cur; 110 next.x=cur.x-1; 111 next.ma[cur.x]=next.ma[next.x]; 112 next.ma[next.x]=9; 113 ans=hsh(next.ma); 114 if(!visit[ans]) 115 { 116 next.g++; 117 next.f=next.g+h(next.ma); 118 visit[ans]=1; 119 next.ans1=ans; 120 pre[ans]=cur.ans1; 121 visit1[ans]='l'; 122 if(ans==ed) return; 123 q.push(next); 124 } 125 } 126 if(cur.x%3!=0)//向右翻 127 { 128 next=cur; 129 next.x=cur.x+1; 130 next.ma[cur.x]=next.ma[next.x]; 131 next.ma[next.x]=9; 132 ans=hsh(next.ma); 133 if(!visit[ans]) 134 { 135 next.g++; 136 next.f=next.g+h(next.ma); 137 visit[ans]=1; 138 next.ans1=ans; 139 pre[ans]=cur.ans1; 140 visit1[ans]='r'; 141 if(ans==ed) return; 142 q.push(next); 143 } 144 } 145 } 146 } 147 int check(int s[]) 148 { 149 int i,j,cnt=0; 150 for(i=1;i<=9;i++) 151 { 152 if(s[i]==9) continue; 153 for(j=1;j<i;j++) 154 { 155 if(s[j]==9) continue; 156 if(s[j]>s[i]) cnt++; 157 } 158 } 159 return cnt; 160 } 161 int main() 162 { 163 int i,j,ans; 164 char str[50]; 165 while(gets(str)) 166 { 167 ans=0; 168 memset(visit,0,sizeof(visit)); 169 for(i=0;str[i];i++) 170 if(str[i]=='x') a[++ans]=9; 171 else if(str[i]!=' ') a[++ans]=str[i]-'0'; 172 ed=0; 173 ans=check(a); 174 if(ans%2) {puts("unsolvable");continue;} 175 bfs(); 176 j=0; 177 while(ed!=st) 178 { 179 str[j++]=visit1[ed]; 180 ed=pre[ed]; 181 } 182 for(i=j-1;i>=0;i--) 183 printf("%c",str[i]); 184 puts(""); 185 } 186 return 0; 187 }