BFS 写的8数码问题 POJ 1077
继8皇后问题后又一大突破8数码问题
Eight
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 8 Accepted Submission(s) : 4
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:
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.
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 a description of a configuration of the 8 puzzle. The 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.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
PKU
整个程序由一下几个部分构成：结点，cantor展开（当HASH函数用的），移动函数，BFS，输出函数
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <stdio.h>
#define MAX 500000
char str[400000];
int tot=0;
using namespace std;
struct Note
{
char s[9];//数码串
int num;//第多少号
int fa;//父结点是多少号
char dir;//从父结点MOVE方向
}note[MAX];
queue<Note> q;
int vis[MAX];
int cantor(char* s)
{
int len=9;
int sum=0;
for(int i=0;i<len;i++)
{
int a=0;
int b=1;
for(int j=i+1;j<len;j++)
if(s[j]<s)
{
a=a+1;
}
//sum+=a*(8-i)!
for(int k=1;k<len-i;k++)
{
b*=k;
}
sum+=a*b;
}
return sum;
}
int move4(char *s,int n)
{
//会对S进行改变
//返回1表示进行了移动，返回0表示没有移动
int pos;
int len=9;
if(n==1)//move up
{
for(int i=0;i<len;i++)
{
if(s=='9')
{
pos=i;
break;
}
}
if(pos-3>=0)
{
s[pos]=s[pos-3]^s[pos];
s[pos-3]=s[pos]^s[pos-3];
s[pos]=s[pos]^s[pos-3];
return 1;
}
else
return 0;
}
if(n==2)//move down
{
for(int i=0;i<len;i++)
{
if(s=='9')
{
pos=i;
break;
}
}
if(pos+3<9)
{
s[pos]=s[pos+3]^s[pos];
s[pos+3]=s[pos]^s[pos+3];
s[pos]=s[pos]^s[pos+3];
return 1;
}
else
return 0;
}
if(n==3)//move left
{
for(int i=0;i<len;i++)
{
if(s=='9')
{
pos=i;
break;
}
}
if(pos-1>=0&&(pos/3==(pos-1)/3))
{
s[pos]=s[pos-1]^s[pos];
s[pos-1]=s[pos]^s[pos-1];
s[pos]=s[pos]^s[pos-1];
return 1;
}
else
return 0;
}
if(n==4)//move right
{
for(int i=0;i<len;i++)
{
if(s=='9')
{
pos=i;
break;
}
}
if(pos+1<9&&(pos/3==(pos+1)/3))
{
s[pos]=s[pos+1]^s[pos];
s[pos+1]=s[pos]^s[pos+1];
s[pos]=s[pos]^s[pos+1];
return 1;
}
else
return 0;
}
return 0;
}
int bfs()
{
while(!q.empty())
{
int ok=1;
Note t=q.front();
Note t2=q.front();
q.pop();
for(int i=0;i<8;i++)
{
if(t.s-48==i+1&&t.s[8]=='9')
;
else
{
ok=0;
}
}
// cout<<"判断是否可以结束: "<<ok<<endl;
if(ok)
{
return t.fa;
}
// cout<<"还要继续搜索。。。"<<endl;
t2=t;//cout<<"up up up\n";
if(move4(t2.s,1))
{
// cout<<"向上走。。。"<<endl;
int k=cantor(t2.s);
if(vis[k]==0)
{
// cout<<"行。。。"<<endl;
t2.dir='u';
t2.fa=t.num;
/*
cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
for(int i=0;i<9;i++)
{
cout<<t2.s<<" ";
if((i+1)%3==0) cout<<endl;
}
cout<<t2.dir<<endl;
*/
t2.num=k;
note[k]=t2;
q.push(t2);
vis[k]=1;
}
// else cout<<"不行。。。"<<endl;
}
t2=t; // cout<<"down down down\n";
if(move4(t2.s,2))
{
// cout<<"向下走。。。"<<endl;
int k=cantor(t2.s);
if(vis[k]!=1)
{
t2.dir='d';
t2.fa=t.num;
// cout<<"行。。。"<<endl;
/*
cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
for(int i=0;i<9;i++)
{
cout<<t2.s<<" ";
if((i+1)%3==0) cout<<endl;
}
cout<<t2.dir<<endl;
*/
t2.num=k;
note[k]=t2;
q.push(t2);
vis[k]=1;
}
// else cout<<"不行。。。"<<endl;
}
t2=t; // cout<<"left left left\n";
if(move4(t2.s,3))
{
// cout<<"向左走。。。"<<endl;
int k=cantor(t2.s);
if(vis[k]!=1)
{
t2.dir='l';
t2.fa=t.num;
// cout<<"行。。。"<<endl;
/*
cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
for(int i=0;i<9;i++)
{
cout<<t2.s<<" ";
if((i+1)%3==0) cout<<endl;
}
cout<<t2.dir<<endl;
*/ t2.num=k;
note[k]=t2;
q.push(t2);
vis[k]=1;
}
// else cout<<"不行。。。"<<endl;
}
t2=t; //cout<<"right right right\n";
if(move4(t2.s,4))
{
// cout<<"向右走。。。"<<endl;
int k=cantor(t2.s);
if(vis[k]!=1)
{
t2.dir='r';
t2.fa=t.num;
// cout<<"行。。。"<<endl;
/*
cout<<"k: "<<k<<" fa:"<<t2.fa<<endl;
for(int i=0;i<9;i++)
{
cout<<t2.s<<" ";
if((i+1)%3==0) cout<<endl;
}
cout<<t2.dir<<endl;
*/
t2.num=k;
note[k]=t2;
q.push(t2);
vis[k]=1;
}
// else cout<<"不行。。。"<<endl;
}
}
return -999;
}
void print (int p)
{
stack < char > stck;
while ( note[p].dir!='Q' )
{
stck.push(note[p].dir);
p = note[p].fa;
}
while ( !stck.empty() )
{
putchar ( stck.top() );
stck.pop();
}
putchar ('\n');
}
int main()
{
memset(vis,0,sizeof(vis));
Note t;
for(int i=0;i<9;i++)
{
char c;
cin>>c;
if(c=='x')
{
t.s='9';
}
else t.s=c;
}
int k=cantor(t.s);
t.num=k;
t.fa=-998;
t.dir='Q';
note[k]=t;
q.push(t);
int B=bfs();
note[k].dir='Q';
note[k].fa=-998;
///xxxxxx
if(B==-999) cout<<"unsolvable"<<endl;
// cout<<"B: "<<B<<endl;
print(0);
return 0;
}
