HDU1043 eight 八数码问题

        八数码问题:

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 1213 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 1213 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, 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 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. Do not print a blank line between cases.

 

Sample Input

2  3  4  1  5  x  7  6  8

 

Sample Output

ullddrurdllurdruldr

本题有多种答案，采用special judge

注意：如何判断给出的序列无解.

      可以序列的逆序对进行剪枝。对于 x，如果在所在行移动，逆序数对不变，如果在列上移动，逆序数对成偶数加减，那么逆序对必须为even。（偶数）

我设置两个队列，采用双向广搜的方法，poj 79ms.

下面附代码：

#include <iostream>#include <map>#include <cstdio>using namespace std;#define M 500000struct state{    int father,code;    char  dir;};state que[M],reque[M];char path[M];// 答案数组void codes(int code){   char s[10];   sprintf(s,"%d",code);    for(int i=0;i<9;i++)    {       if(i!=0&& i%3==0)         printf("\n");       printf("%c ",s[i]);    }     printf("\n\n");}int main(){    int dir[4][2]= {{-1,0},{1,0},{0,-1},{0,1}}; //方向数组    char str[4]={'u','d','l','r'};    char str1[4]={'d','u','r','l'};    int base[9]= {100000000,10000000,1000000,100000,10000,1000,100,10,1};    map<int,int> a;    char w[3][3],w1[9];    int i,j,counter,qhead,qrear,re_qhead,re_qrear,ji,ji1,x,y,k;    int newx,newy;   while(cin>>w[0][0]){cin>>w[0][1]>>w[0][2];w1[0]=w[0][0];w1[1]=w[0][1];w1[2]=w[0][2];for(i=1;i<3;i++)for(j=0;j<3;j++){   cin>>w[i][j];            w1[i*3+j]=w[i][j];// convert to one array}    //求逆序对数    counter=0;    for(i=0; i<8; ++i)        for(j=i+1; j<9; ++j)            if(w1[i]!='x'&&w1[j]!='x'&&w1[i]>w1[j])                counter++;    //如果逆序对数是 奇数，那么无法转化为目标状态    if(counter&1)    {        printf("unsolveble!\n");    }    else    {        int e,s;        s=0;        e=123456789;//反向搜索的起始状态        //将八数码转换为一个整数s;        for(i=0; i<9; i++)        {            if(w1[i]=='x')                s=s*10+9;            else                s=s*10+(w1[i]-'0');        }        //init        qhead=qrear=re_qhead=re_qrear=1;        que[qrear].code=s;    //队列1        reque[re_qrear].code=e;//队列2        a[s]=1;//正向搜索 已访问 标记为1        a[e]=2;//反向搜索 已访问 标记为2        bool flag=0;        if(s==e)//特殊情况处理            flag=1;        while(!flag)        {            for(ji=0; ji<9; ji++) //提取正着搜时x所在的位置                if((que[qhead].code/base[ji])%10==9)                    break;            for(ji1=0; ji1<9; ji1++) //提取反着搜时x所在的位置                if((reque[re_qhead].code/base[ji1])%10==9)                    break;            for(i=0; i<4; i++) //双向广搜            {                x=ji/3;                y=ji%3;                newx=x+dir[i][0];                newy=y+dir[i][1];                if(newx<3&&newx>=0&&newy<3&&newy>=0&&flag==0)                {                    s=que[qhead].code;//s为当前的code                    //求新的code                    int d1=9;//d1=(s/base[ji])%10;//可以直接赋值为9                    int d2=(s/base[newx*3+newy])%10;//与之交换的数放在d2中                    s=s-d1*base[ji] - d2*base[newx*3+newy];                    s=s+d1*base[newx*3 + newy]+d2*base[ji];                    if(a[s]!=1)                    {                        if(a[s]==2)                            flag=1;//搜到了                        a[s]=1;//标记                        qrear++;                        que[qrear].code=s;                        que[qrear].father=qhead;//记录该节点是由哪个节点扩展来的                        que[qrear].dir=str[i];//标记方向                    }                }                x=ji1/3;                y=ji1%3;                newx=x+dir[i][0];                newy=y+dir[i][1];                if(newx<3&&newx>=0&&newy<3&&newy>=0&&flag==0)                {                    s=reque[re_qhead].code;//找到新的S                    //求新的code                    int d1=9;//d1=(s/base[ji1])%10;//可以直接赋值为9                    int d2=(s/base[newx*3+newy])%10;//与之交换的数放在d2中                    s=s-d1*base[ji1] - d2*base[newx*3+newy];                    s=s+d1*base[newx*3 + newy]+d2*base[ji1];                    if(a[s]==0)                    {                        a[s]=2;//标记                        re_qrear++;                        reque[re_qrear].code=s;                        reque[re_qrear].father=re_qhead;//记录该节点是由哪个节点扩展来的                        reque[re_qrear].dir=str1[i];//标记方向                    }                }            }            qhead++;//从队列1中取下一个节点            re_qhead++;//从队列2 中取下一个节点        }//end while        //下面提取路径        counter=0;        k=qrear;        //printf("%d\n",que[qrear].code);        //codes(que[qrear].code);        while(k!=1)        {           path[counter++]=que[k].dir;           k=que[k].father;        }        for(i=counter-1;i>=0;i--)         printf("%c",path[i]);      // printf("\nup is first\n");        k=1;counter=0;        while(k<=re_qrear&&reque[k].code!=que[qrear].code)        {           k++;        }        //printf("%d\n",reque[reque[k].father].code);        while(k!=1)        {           path[counter++]=reque[k].dir;           k=reque[k].father;         //  codes(reque[k].code);        }        for(i=0;i<counter;i++)          printf("%c",path[i]);        printf("\n");        a.clear();    }}   return 0;}/**2 3 41 5 x7 6 81 2 34 5 67 8 x*/