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: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.
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 puzzle1 2 3 x 4 6 7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
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.
2 3 4 1 5 x 7 6 8
ullddrurdllurdruldr
这是一个八数码问题,关于这道题的题意,可以参考Virtual Judge
上面的翻译,我们的思路是从初始状态开始搜索,然后搜索完所有的状态,每次搜索的时候记录路径,并且采用康托展开进行判重,因为我们已经预处理好了所有的状态。所以最后应对 O(1) O ( 1 ) 的查询就好了
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
const int N=4e5+10;
int fac[10]= {0,1,2,6,24,120,720,5040,40320,362880};
string path[N];
int vis[N];
int go[4][2]= {-1,0,1,0,0,-1,0,1}; //上下左右
char dir[]="durl";//和方向变量相反
struct node
{
int s[9];//序列
int zero_pos;//0的位置
int hash;//哈希值
string path;//路径
};
int cantor_hash(int s[])//康托展开判重
{
int res=0;
for(int i=0; i<9; i++)
{
int num=0;
for(int j=i+1; j<9; j++)
if(s[i]>s[j])
num++;
res+=(num*fac[9-i-1]);
}
return res;
}
void init()
{
node now,to;
mem(vis,0);
for(int i=0; i<8; i++)
now.s[i]=i+1;
now.s[8]=0;
now.zero_pos=8;
now.hash=cantor_hash(now.s);
path[now.hash]="";
vis[now.hash]=1;
queue q;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
int x=now.zero_pos/3;
int y=now.zero_pos%3;
for(int i=0; i<4; i++)
{
int xx=x+go[i][0];
int yy=y+go[i][1];
if(xx>=0&&xx<3&&yy>=0&&yy<3)
{
to=now;
to.zero_pos=xx*3+yy;
to.s[now.zero_pos]=to.s[to.zero_pos];
to.s[to.zero_pos]=0;
to.hash=cantor_hash(to.s);
if(!vis[to.hash])
{
to.path=dir[i]+now.path;
path[to.hash]=to.path;
q.push(to);
vis[to.hash]=1;
}
}
}
}
}
int main()
{
init();
node ans;
char s[5];
while(~scanf("%s",s))
{
if(s[0]=='x')
{
ans.zero_pos=0;
ans.s[0]=0;
}
else
ans.s[0]=s[0]-'0';
for(int i=1; i<9; i++)
{
scanf("%s",s);
if(s[0]=='x')
{
ans.zero_pos=i;
ans.s[i]=0;
}
else
ans.s[i]=s[0]-'0';
}
ans.hash=cantor_hash(ans.s);
if(vis[ans.hash])
cout<else
puts("unsolvable");
}
return 0;
}