POJ 1077八数码问题（cantor展开+BFS）

Eight

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28887 Accepted: 12584 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

广度搜索很好想，主要难想的地方在hash判重，这里用到的是cantor展开判重复

基本思想是，从全排列组成的数字中，找出比他小的有数有多少

对于一个数字，从第一位看起，找他后面有多少个比这一位小的位，比如有3个，那就是假设如果交换

这两位那么后面的数无论怎么排列都比以前小，也就是比原来那个数小的数有剩下位数全排列种类数的个数。

比如213456789

第一位2后面有一个数1比他小，假设交换

1（23456789）括号里面的数无论怎么排列都比原来的213456789小吧

那就是8！种

再比如312456789

第一位3后面有两个数1和2比他小，假设交换

1（32456789）括号里面的数无论怎么排列都比原来的312456789小吧

2（13456789）括号里面的数无论怎么排列都比原来的312456789小吧

那就是8！+8！种

其他位数也是都是这样运算的。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<queue>using namespace std;struct node {    int a[3][3] ;    int x,y ;    char states[100];    node(){        memset(states,'\0',sizeof(states));        memset(a,0,sizeof(a));        x=0;y=0;    }};node h;const int MAXN=1000000;///最多是9!int fac[]={1,1,2,6,24,120,720,5040,40320,362880};///康拖展开判重0!1!2!3!4!5!6!7!8!9!bool vis[MAXN];///标记int addx[4]={-1,1,0,0};int addy[4]={0,0,-1,1};int cantor(int m[3][3])///康拖展开求该序列的hash值{    int s[9];    int k=0;    for(int i=0;i<3;i++){        for(int j=0;j<3;j++){            s[k++]=m[i][j];        }    }    int sum=0;    for(int i=0;i<9;i++)    {        int num=0;        for(int j=i+1;j<9;j++)          if(s[j]<s[i])num++;        sum+=(num*fac[9-i-1]);    }    return sum+1;}void bfs(){    queue<node> qq;    qq.push(h);    while(!qq.empty()){        node top=qq.front();        qq.pop();        if(cantor(top.a)==1){            printf("%s\n",top.states);            return ;        }        int ch[3][3];        for(int i=0;i<3;i++){            for(int j=0;j<3;j++){                ch[i][j]=top.a[i][j];            }        }        for(int i=0;i<4;i++){            int newx=top.x+addx[i];            int newy=top.y+addy[i];            if(newx>=0&&newx<3&&newy>=0&&newy<3){                swap(ch[newx][newy],ch[top.x][top.y]);                if(!vis[cantor(ch)]){                    node pp;                    memcpy(pp.a,ch,9*sizeof(int));///快速复制数组                    memcpy(pp.states,top.states,sizeof(top.states));                    pp.x=newx;                    pp.y=newy;                    if(i==0)strcat(pp.states,"u");                    if(i==1)strcat(pp.states,"d");                    if(i==2)strcat(pp.states,"l");                    if(i==3)strcat(pp.states,"r");                    qq.push(pp);                    vis[cantor(pp.a)]=true;                }                swap(ch[newx][newy],ch[top.x][top.y]);            }        }    }}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int arry[9];    for(int i=0;i<9;i++){        char c[2];        scanf("%s",&c);        if(c[0]=='x')arry[i]=9;        else arry[i]=c[0]-'0';    }    int k=0;    for(int i=0;i<3;i++){        for(int j=0;j<3;j++){            h.a[i][j]=arry[k++];            if(h.a[i][j]==9){                h.x=i;                h.y=j;            }        }    }    memset(vis,false,sizeof(vis));    vis[cantor(h.a)]=true;    bfs();   return 0;}