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
八数码问题，目标状态固定，给数据用的是一维。
A*算法+二叉堆。估价函数写的是位移差的和。
#include <stdio.h>
#include <string.h>
#include <math.h>
struct Eight {
int value;
int status[3][3];
char walked[60];
char str[15];
int xx,xy;
};
struct EightExtented {
bool left,right;
};
EightExtented eighte[370000];
Eight eight[370000];
bool hashtable[363000]={false};
bool solved=false;
int jc[10]={1,1,2,6,24,120,720,5040,40320,362880};
int tail=0;
char start[10];
int hmove[4][2]={{0,1},{0,-1},{-1,0},{1,0}};
int cvalue();
void init()
{
int i;
char temp;
for (i=0;i<9;i++)
{
do
{
scanf("%c",&temp);
}while (temp==' ');
start[i]=temp;
if (start[i]!='x') eight[0].status[i/3][i%3]=start[i]-'0';
else {
eight[0].status[i/3][i%3]=9;
eight[0].xx=i/3;eight[0].xy=i%3;
}
}
eight[0].value=cvalue();
tail++;
}
int hash()
{
int rev[10]={0};
int i,j;
int loc;
int hashvalue=0;
for (i=0;i<9;i++)
{
if (eight[tail].str[i]=='x') {
loc=i;
continue;
}
for (j=i+1;j<9;j++)
{
if ((eight[tail].str[j]!='x')&&(eight[tail].str[i]>eight[tail].str[j])) rev[eight[tail].str[i]-'0'-1]++;
}
}
for (i=0;i<9;i++) hashvalue+=rev[i]*jc[i];
hashvalue+=jc[8]*(9-loc-1);
return hashvalue;
}
bool able()
{
int i,j;
int sum=0;
for (i=0;i<8;i++)
for (j=i+1;j<9;j++)
if ((start[i]!='x')&&(start[j]!='x')&&(start[i]>start[j])) sum++;
if (sum%2==0) return true;
else return false;
}
int cvalue()
{
int i,j;
int value=0;
for (i=0;i<3;i++)
for (j=0;j<3;j++)
value+=fabs((eight[tail].status[i][j]-1)/3-i)+fabs((eight[tail].status[i][j]-1)%3-j);
return value;
}
//0 right, 1 left, 2 up, 3 down
bool move(int way)
{
int x,y;
int i,j,l;
char temp;
if (strlen(eight[0].walked)>=50) {
return false;
}
x=eight[0].xx+hmove[way][0];
y=eight[0].xy+hmove[way][1];
if (!((x>=0)&&(x<3)&&(y>=0)&&(y<3))) return false;
for (i=0;i<3;i++)
for (j=0;j<3;j++) eight[tail].status[i][j]=eight[0].status[i][j];
eight[tail].status[eight[0].xx][eight[0].xy]=eight[tail].status[x][y];
eight[tail].status[x][y]=9;
eight[tail].xx=x;eight[tail].xy=y;
for (i=0;i<3;i++)
for (j=0;j<3;j++) if (eight[tail].status[i][j]!=9) eight[tail].str[i*3+j]=eight[tail].status[i][j]+'0';
else eight[tail].str[i*3+j]='x';
eight[tail].str[9]=0;
if (hashtable[hash()]) return false;
else hashtable[hash()]=true;
strcpy(eight[tail].walked,eight[0].walked);
switch (way)
{
case 0: temp='r';break;
case 1: temp='l';break;
case 2: temp='u';break;
case 3: temp='d';break;
}
l=strlen(eight[tail].walked);
eight[tail].walked[l]=temp;
eighte[tail].left=-1;eighte[tail].right=-1;
eight[tail].walked[l+1]=0;
eight[tail].value=cvalue()*10+l+1;
eighte[tail].left=false;eighte[tail].right=false;
return true;
}
void solve()
{
int i,t;
Eight temp;
while (!solved)
{
for (i=0;i<4;i++)
{
if (move(i)) {
if (tail>1) {
t=tail;
while ((t>1)&&(eight[t].value<eight[t/2].value))
{
temp=eight[t/2];
eight[t/2]=eight[t];
eight[t]=temp;
t/=2;
}
}
if (tail%2==0) eighte[tail/2].left=true;
else eighte[tail/2].right=true;
tail++;
}
}
if (eight[0].value-strlen(eight[0].walked)==0) {
printf("%s\n",eight[0].walked);
solved=true;
}
eight[0]=eight[1];
eight[1]=eight[tail-1];
//printf("%s %s %d\n",eight[0].str,eight[0].walked,eight[0].value);
tail--;
if (tail%2==0) eighte[tail/2].left=false;
else eighte[tail/2].right=false;
t=1;
while (t*2<tail)
{
if (eight[t].value>eight[t*2].value) {
if (eighte[t*2+1].right&&eight[t*2+1].value<eight[t*2].value) {
temp=eight[t];
eight[t]=eight[t*2+1];
eight[t*2+1]=temp;
t=t*2+1;
}
else {
temp=eight[t];
eight[t]=eight[t*2];
eight[t*2]=temp;
t=t*2;
}
}
else if (eighte[t*2+1].right&&eight[t].value>eight[t*2+1].value) {
temp=eight[t];
eight[t]=eight[t*2+1];
eight[t*2+1]=temp;
t=t*2+1;
}
else break;
}
}
}
int main()
{
//freopen("test.out","w",stdout);
init();
if (!able()) {
printf("unsolvable\n");
return 0;
}
else solve();
return 0;
}