HDU 1043 - Eight

A - Eight

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1043

Appoint description: System Crawler  (2016-01-16)

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 

用康托展开做Hash，用BFS从最终状态倒推，得到每种状态的最短路径，路径用后驱的康托数值和对应的操作来记录。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>using namespace std;#define N 363000struct node{    int a[10];    int pos;    int cant;    node(int *b,int p,int c){        memcpy(a,b,sizeof(a));        pos=p;        cant=c;    }};int vx[]={0,0,1,-1};int vy[]={1,-1,0,0};char op[]="lrud";int pre[N];bool vis[N];char cz[N];int fac[]={1,1,2,6,24,120,720,5040,40320,362880};int Cantor(int *s,int n){    int num=0;    for (int i=0;i<n;i++){        int cnt=0;        for (int j=i+1;j<n;j++)            cnt+=(s[j]<s[i]);        num+=fac[n-i-1]*cnt;    }    return num;}void bfs(){    memset(pre,-1,sizeof(pre));    memset(vis,false,sizeof(vis));    int a[]={1,2,3,4,5,6,7,8,0};    int cant=Cantor(a,9);    int pos=8;    queue<node> q;    vis[cant]=true;    q.push(node(a,pos,cant));    while(!q.empty()){        node temp=q.front();        q.pop();        int x=temp.pos/3;        int y=temp.pos%3;        for (int i=0;i<4;i++){            int tx=x+vx[i];            int ty=y+vy[i];            if (tx<0 || tx>=3 || ty<0 || ty>=3)                continue;            memcpy(a,temp.a,sizeof(a));            swap(a[temp.pos],a[tx*3+ty]);            cant=(Cantor(a,9));            if (!vis[cant]){                vis[cant]=true;                pre[cant]=temp.cant;                cz[cant]=op[i];                q.push(node(a,tx*3+ty,cant));            }        }    }}void pr(int cant){    if (pre[cant]==-1)        return;    printf("%c",cz[cant]);    pr(pre[cant]);}int main(){    bfs();    char st[50];    while (gets(st)!=NULL){        int s[10],k=0;        memset(s,-1,sizeof(s));        for (int len=strlen(st),i=0;i<len;i++){            if (st[i]=='x'){                s[k++]=0;              //  cout<<s[k-1];            }            else{                if (st[i]!=' '){                    s[k++]=st[i]-'0';                    //cout<<s[k-1];                }            }        }       // cout<<endl;        int cant=Cantor(s,9);        if (!vis[cant])            printf("unsolvable");        else            pr(cant);        printf("\n");    }    return 0;}