http://acm.hdu.edu.cn/showproblem.php?pid=1043
Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30907 Accepted Submission(s): 8122
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
题解：裸的康拓展开
使用康拓展开对全排列进行hash，然后使用bfs从末状态开始跑，跑到的状态都是有解的状态，否则输入的初状态无解，在bfs过程记录这一步的下一步即可。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 using namespace std; 6 int qwq[10]; 7 char sss[403200]; 8 int vv[403200]; 9 bool v[403200]; 10 int f0; 11 int counter(string s){ 12 int sum=0; 13 for(int i=2;i<=8;i++){ 14 qwq[i]=qwq[i-1]*i; 15 } 16 for(int i=0;i<9;i++){ 17 int tt=0; 18 for(int j=i+1;j<9;j++){ 19 if(s[j]<s[i]){ 20 tt++; 21 } 22 } 23 sum+=qwq[8-i]*tt; 24 } 25 return sum; 26 } 27 queue<string>pq; 28 void bfs(){ 29 while(!pq.empty())pq.pop(); 30 string s0="123456789"; 31 pq.push(s0); 32 f0=counter(s0); 33 v[f0]=1; 34 while(!pq.empty()){ 35 string aa=pq.front();pq.pop(); 36 int fs=counter(aa); 37 //if(fs==76346)cout <<aa<<endl; 38 for(int i=0;i<aa.size();i++){ 39 if(aa[i]=='9'){ 40 if(i%3==0){ 41 string bb=aa; 42 bb[i]=bb[i+1]; 43 bb[i+1]='9'; 44 int f=counter(bb); 45 if(!v[f]){ 46 v[f]=1; 47 sss[f]='l'; 48 vv[f]=fs; 49 pq.push(bb); 50 } 51 if(i!=6){ 52 string cc=aa; 53 cc[i]=cc[i+3]; 54 cc[i+3]='9'; 55 int f=counter(cc); 56 if(!v[f]){ 57 v[f]=1; 58 vv[f]=fs; 59 sss[f]='u'; 60 pq.push(cc); 61 } 62 } 63 if(i!=0){ 64 string dd=aa; 65 dd[i]=dd[i-3]; 66 dd[i-3]='9'; 67 int f=counter(dd); 68 if(!v[f]){ 69 v[f]=1; 70 vv[f]=fs; 71 sss[f]='d'; 72 pq.push(dd); 73 } 74 } 75 } 76 else if(i%3==1){ 77 string bb=aa; 78 bb[i]=bb[i+1]; 79 bb[i+1]='9'; 80 int f=counter(bb); 81 if(!v[f]){ 82 v[f]=1; 83 vv[f]=fs; 84 sss[f]='l'; 85 pq.push(bb); 86 } 87 string cc=aa; 88 cc[i]=cc[i-1]; 89 cc[i-1]='9'; 90 f=counter(cc); 91 if(!v[f]){ 92 v[f]=1; 93 vv[f]=fs; 94 sss[f]='r'; 95 pq.push(cc); 96 } 97 if(i!=7){ 98 string cc=aa; 99 cc[i]=cc[i+3]; 100 cc[i+3]='9'; 101 int f=counter(cc); 102 if(!v[f]){ 103 v[f]=1; 104 vv[f]=fs; 105 sss[f]='u'; 106 pq.push(cc); 107 } 108 109 } 110 if(i!=1){ 111 string dd=aa; 112 dd[i]=dd[i-3]; 113 dd[i-3]='9'; 114 int f=counter(dd); 115 if(!v[f]){ 116 v[f]=1; 117 vv[f]=fs; 118 sss[f]='d'; 119 pq.push(dd); 120 } 121 } 122 } 123 else{ 124 string bb=aa; 125 bb[i]=bb[i-1]; 126 bb[i-1]='9'; 127 128 int f=counter(bb); 129 // if(fs==76346) 130 //cout <<f<<" "<<endl; 131 if(!v[f]){ 132 v[f]=1; 133 vv[f]=fs; 134 sss[f]='r'; 135 pq.push(bb); 136 } 137 if(i!=8){ 138 string cc=aa; 139 cc[i]=cc[i+3]; 140 cc[i+3]='9'; 141 int f=counter(cc); 142 143 if(!v[f]){ 144 v[f]=1; 145 vv[f]=fs; 146 sss[f]='u'; 147 pq.push(cc); 148 } 149 } 150 if(i!=2){ 151 string dd=aa; 152 dd[i]=dd[i-3]; 153 dd[i-3]='9'; 154 int f=counter(dd); 155 if(!v[f]){ 156 v[f]=1; 157 vv[f]=fs; 158 sss[f]='d'; 159 pq.push(dd); 160 } 161 } 162 } 163 } 164 } 165 } 166 } 167 void init(){ 168 qwq[0]=qwq[1]=1; 169 for(int i=2;i<=8;i++){ 170 qwq[i]=qwq[i-1]*i; 171 } 172 } 173 int main(){ 174 init(); 175 bfs(); 176 char aa[100]; 177 char bb[100]; 178 //while(scanf("%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c",aa[0],aa[1],aa[2],aa[3],aa[4],aa[5], 179 // aa[6],&aa[7],&aa[8],)){ 180 while(gets(aa)){ 181 if(aa[0]==' ')break; 182 int tot=0; 183 int len=strlen(aa); 184 for(int i=0;i<len;i++){ 185 if((aa[i]>='0'&&aa[i]<='8')){ 186 bb[tot++]=aa[i]; 187 } 188 if(aa[i]=='x'){ 189 bb[tot++]='9'; 190 } 191 } 192 bb[tot]=' '; 193 int f=counter(bb); 194 // cout << f<<" "<<bb<<endl; 195 //cout <<f<<endl;46103 196 if(!v[f]){ 197 printf("unsolvable "); 198 } 199 else{ 200 for(int i = f ; i != f0 ; i=vv[i]){ 201 // cout << vv[<<endl; 202 printf("%c",sss[i]); 203 //system("pause"); 204 } 205 printf(" "); 206 } 207 } 208 return 0; 209 }