http://acm.pku.edu.cn/JudgeOnline/problem?id=1077
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.
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
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
South Central USA 1998八數(shù)碼 經(jīng)典的BFS啊， 我就直接用的單向的BFS，也過了，雙向的應(yīng)該會快很多。
判重的hash函數(shù)看的discuss里的，用逆序數(shù)，9!種情況，一一對應(yīng)，不會有重復(fù)的。
Source Code
| Problem: 1077 |
|
User: lovecanon |
| Memory: 9572K |
|
Time: 125MS |
| Language: GCC |
|
Result: Accepted |
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct node
{
int state[3][3];
int pre;
int dir;
}queue[362881];
int hash[362881];
int step[362881];
int a[3][3];
int fac(int i)
{
switch(i)
{
case 0: return 1;
case 1: return 1;
case 2: return 2;
case 3: return 6;
case 4: return 24;
case 5: return 120;
case 6: return 720;
case 7: return 5040;
case 8: return 40320;
}
return 0;
}
int HASH()
{
int i,j,k=0,b[9],ret=0,num=0;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
b[k++]=a[i][j];
for(i=0;i<9;i++)
{
num=0;
for(j=0;j<i;j++)
if(b[j]>b[i]) num++;
ret+=fac(i)*num;
}
return ret;
}
void output(int len)
{
int i;
for(i=len;i>=0;i--)
{
if(step[i]==1) printf("l");
if(step[i]==2) printf("r");
if(step[i]==3) printf("u");
if(step[i]==4) printf("d");
}
printf("\n");
}
int main()
{
char s[10];
int i,j,rear=0,front=0,tag=0;
rear++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
{
scanf("%s",s);
if(s[0]=='x') s[0]='9';
queue[rear].state[i][j]=s[0]-'0';
}
queue[rear].pre=0;queue[rear].dir=0;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
a[i][j]=queue[rear].state[i][j];
hash[HASH()]=1;
while(front<rear)
{
int e,f,tmp,cntdir,len;
front++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
{
a[i][j]=queue[front].state[i][j];
if(a[i][j]==9) {e=i;f=j;}
}
if(f-1>=0)
{
cntdir=1;
tmp=a[e][f];
a[e][f]=a[e][f-1];
a[e][f-1]=tmp;
tmp=HASH();
if(tmp==0)
{
int t=front;
len=0;
step[len++]=cntdir;
while(queue[t].pre)
{
step[len++]=queue[t].dir;
t=queue[t].pre;
}
output(len);
return 0;
}
if(!hash[tmp])
{
rear++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
queue[rear].state[i][j]=a[i][j];
queue[rear].dir=cntdir;queue[rear].pre=front;
hash[tmp]=1;
}
tmp=a[e][f];
a[e][f]=a[e][f-1];
a[e][f-1]=tmp;
}
if(f+1<3)
{
cntdir=2;
tmp=a[e][f];
a[e][f]=a[e][f+1];
a[e][f+1]=tmp;
tmp=HASH();
if(tmp==0)
{
int t=front;
len=0;
step[len++]=cntdir;
while(queue[t].pre)
{
step[len++]=queue[t].dir;
t=queue[t].pre;
}
output(len);
return 0;
}
if(!hash[tmp])
{
rear++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
queue[rear].state[i][j]=a[i][j];
queue[rear].dir=cntdir;queue[rear].pre=front;
hash[tmp]=1;
}
tmp=a[e][f];
a[e][f]=a[e][f+1];
a[e][f+1]=tmp;
}
if(e-1>=0)
{
cntdir=3;
tmp=a[e][f];
a[e][f]=a[e-1][f];
a[e-1][f]=tmp;
tmp=HASH();
if(tmp==0)
{
int t=front;
len=0;
step[len++]=cntdir;
while(queue[t].pre)
{
step[len++]=queue[t].dir;
t=queue[t].pre;
}
output(len);
return 0;
}
if(!hash[tmp])
{
rear++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
queue[rear].state[i][j]=a[i][j];
queue[rear].dir=cntdir;queue[rear].pre=front;
hash[tmp]=1;
}
tmp=a[e][f];
a[e][f]=a[e-1][f];
a[e-1][f]=tmp;
}
if(e+1<3)
{
cntdir=4;
tmp=a[e+1][f];
a[e+1][f]=a[e][f];
a[e][f]=tmp;
tmp=HASH();
if(tmp==0)
{
int t=front;
len=0;
step[len++]=cntdir;
while(queue[t].pre)
{
step[len++]=queue[t].dir;
t=queue[t].pre;
}
output(len);
return 0;
}
if(!hash[tmp])
{
rear++;
for(i=0;i<3;i++)
for(j=0;j<3;j++)
queue[rear].state[i][j]=a[i][j];
queue[rear].dir=cntdir;queue[rear].pre=front;
hash[tmp]=1;
}
tmp=a[e+1][f];
a[e+1][f]=a[e][f];
a[e][f]=tmp;
}
}
printf("unsolvable\n");
return 0;
}