八數位(A* 735ms)

20091006

比較經典的一個題目，下面是原題和My Code。

1077 Eight

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6197		Accepted: 2670		Special Judge

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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

typedef struct node
{
    char sta[3][3];
    char posr,posc;
    char ope;
    int lev;
    int dis;
    struct node *next;
}NODE;

int Hash(NODE *cur)
{
    static int i,j,sum,pro,rev;

    sum=0;

    for(pro=i=1;i<9;i++)
    {
        pro*=i;

        for(rev=j=0;j<i;j++)
            if(cur->sta[0][j]>cur->sta[0][i])
                rev++;

        sum+=pro*rev;
    }

    return sum;
}

void Change(NODE *cur,char ope)
{
    char t;

    t=cur->sta[cur->posr][cur->posc];

    switch(ope)
    {
        case 'u':
            cur->sta[cur->posr][cur->posc]=cur->sta[cur->posr+1][cur->posc];
            cur->posr++;

            break;
        case 'd':
            cur->sta[cur->posr][cur->posc]=cur->sta[cur->posr-1][cur->posc];
            cur->posr--;

            break;
        case 'l':
            cur->sta[cur->posr][cur->posc]=cur->sta[cur->posr][cur->posc+1];
            cur->posc++;

            break;
        case 'r':
            cur->sta[cur->posr][cur->posc]=cur->sta[cur->posr][cur->posc-1];
            cur->posc--;

            break;
    }

    cur->sta[cur->posr][cur->posc]=t;
}

void PrintPre(NODE *cur,char table[])
{
    char t;

    if((t=table[Hash(cur)])!='\b')
    {
        Change(cur,t);
        PrintPre(cur,table);
        putchar(t);
    }
}

void End(NODE *cur,char table[])
{
    Change(cur,cur->ope);
    PrintPre(cur,table);
    putchar(cur->ope);
}

int GetPar(NODE *cur)
{
    int i,j,sum=0;

    for(i=1;i<9;i++)
        for(j=0;j<i;j++)
            if((*cur->sta)[j]!='9'&&(*cur->sta)[j]>(*cur->sta)[i])
                sum++;

    return sum%2;
}

void BrushDis(NODE *cur)
{
    static int sum,i,j,t;

    for(sum=i=0;i<3;i++)
        for(j=0;j<3;j++)
        {
            t=cur->sta[i][j]-'0'-1;
            sum+=abs(t/3-i)+abs(t%3-j);
        }

    cur->dis=sum;
}

void Insert(NODE *cur,NODE *head)
{
    cur->next=head->next;
    head->next=cur;
}

int main(void)
{
    int i,j,min,r,c;
    char table[362880]={0},temp[20],dir[4][3]={1,0,'d',-1,0,'u',0,1,'r',0,-1,'l'};
    NODE *head,*cur,*t,*pre;

    head=(NODE *)malloc(sizeof(NODE));
    head->next=NULL;

    cur=(NODE *)malloc(sizeof(NODE));
    cur->ope='\b';
    cur->lev=0;

    for(i=0;i<3;i++)
        for(j=0;j<3;j++)
        {
            scanf("%s",temp);
           
            if('x'==*temp)
            {
                cur->posr=i;
                cur->posc=j;
                cur->sta[i][j]='9';
            }
            else
                cur->sta[i][j]=*temp;
        }

    if(GetPar(cur))
    {
        printf("unsolvable");

        return 0;
    }

    BrushDis(cur);
    Insert(cur,head);

    while(1)
    {
        for(min=((unsigned int)(~0))>>1,pre=head,cur=head->next;cur!=NULL;pre=cur,cur=cur->next)
            if(cur->dis+cur->lev<min)
            {
                min=cur->dis+cur->lev;
                t=pre;
            }

        cur=t->next;
        t->next=cur->next;

        if(0==cur->dis)
        {
            End(cur,table);
            free(cur);

            while(head!=NULL)
            {
                pre=head;
                head=head->next;
                free(pre);
            }

            return 0;
        }

        for(i=0;i<4;i++)
        {
            r=cur->posr+dir[i][0];
            c=cur->posc+dir[i][1];

            if(r>=0&&r<3&&c>=0&&c<3)
            {
                t=(NODE *)malloc(sizeof(NODE));

                for(j=0;j<9;j++)
                    *(*t->sta+j)=*(*cur->sta+j);

                t->sta

[c]='9';
                t->sta[cur->posr][cur->posc]=cur->sta

[c];

                if(table[Hash(t)])
                {
                    free(t);

                    continue;
                }

                t->ope=dir[i][2];
                t->posr=r;
                t->posc=c;
                t->lev=cur->lev+1;

                BrushDis(t); //也可用cur->dis計算
                Insert(t,head);
            }
        }

        table[Hash(cur)]=cur->ope;
        free(cur);
    }
}