HDU1043：Eigth(康托展开)

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.

ullddrurdllurdruldr

8数码问题网上解说很多，我是用康托展开做的

#include
#include
#include
#include
#include
using namespace std;
int s[20],pos[20],end[20];
int hash[20],vis[1000000];
string ans[1000000];
struct node
{
string step;
int num[10],val;
int pp;
};
int solve(int s[])
{
int sum=0;
for(int i=0; i<9; i++)
{
int num=0;
for(int j=i+1; j<9; j++)
if(s[j] sum+=(num*hash[9-i-1]);
}
return sum+1;
}
int to[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
char path[10] = "durl";
void bfs()
{
memset(vis,0,sizeof(vis));
int i,j;
node a,next;
queue Q;
for(i = 0; i<8; i++)
a.num[i] = i+1;
a.num[8] = 0;
a.pp = 8;
a.step = "";
a.val = 46234;
ans[a.val] = "";
Q.push(a);
while(!Q.empty())
{
a = Q.front();
Q.pop();
int x = a.pp/3;
int y = a.pp%3;
for(i = 0; i<4; i++)
{
int tx = x+to[i][0];
int ty = y+to[i][1];
if(tx<0 || ty<0 || tx>2 || ty>2)
continue;
next = a;
next.pp = 3*tx+ty;
next.num[a.pp] = next.num[next.pp];
next.num[next.pp] = 0;
next.val = solve(next.num);
if(!vis[next.val])
{
vis[next.val]=true;
next.step=path[i]+next.step;
Q.push(next);
ans[next.val]=next.step;
}
}
}
}
int main()
{
int i;
char ch;
node a;
int c[10];
hash[0] = 1;
for(i = 1; i<=10; i++)
hash[i] = hash[i-1]*i;
bfs();
while(cin >> ch)
{
if(ch == ‘x‘)
c[0] = 0;
else c[0]=ch-‘0‘;
for(int i=1; i<9; i++)
{
cin>>ch;
if(ch==‘x‘)
c[i]=0;
else c[i]=ch-‘0‘;
}
int k;
k = solve(c);
if(vis[k])
cout < else
cout <<"unsolvable" < }
return 0;
}