HUD 1043 Eight

EightTime limit:10000/5000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 14009 Accepted Submission (s): 3965
Special Judge

Problem DescriptionThe 15-puzzle have been around for over years; Even if you don't know it by so name, you ' ve seen it. It is constructed with the sliding tiles, each with a number from 1 to the IT, and all packed into a 4 by 4 frame with on E tile missing. Let's call the missing tile ' x '; The object of the puzzle is to arrange, the tiles so, they are ordered as:

1  2  3  4 5  6  7  8 9 1213  X

Where the only legal operation are 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     9 x  9 (     9)  1213            -ten-ten-----X r->            d->            r->

The letters in the previous row indicate which neighbor of the ' x ' tile are swapped with the ' X ' tiles at each step; Legal values is ' r ', ' l ', ' u ' and ' d ', for-right, left-up, and-down, respectively.

Not all puzzles can solved; In 1870, a man named Sam Loyd is famous for distributing an unsolvable version of the puzzle, and
Frustrating many people. In fact, any of the regular puzzle into a unsolvable one are to swap the tiles (not counting the missing ' X ' tile, of course).

In this problem, you'll write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.

Inputyou would 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 T Iles listed from left to right within a row, where the tiles is 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

Outputyou would print to standard output either the word "unsolvable", if the puzzle has no solution, or a string consist ing 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. Don't print a blank line between cases.

Sample Input

2  3  4  1  5  x  7  6  8

Sample Output

Ullddrurdllurdruldr

Eight digital problems.

Bfs+ Kang Expand Open

#include <iostream> #include <stdio.h> #include <string> #include <cstring> #include < algorithm> #include <cmath> #include <queue> #include <stack> #define N 400000using namespace std;    struct node{int pos,val;    Char a[9]; string path;} Cur,now;int f[15]= {40320,5040,720,120,24,6,2,1,1}; Kang Drag expand the sentence weight int dx[4]={-1,1,0,0};//up and down int dy[4]={0,0,-1,1};char dir[5]= "Durl"; string Anspath[n],ans;int Vis[n];char C;    Queue<node>q;int Kanto (char *ss) {int ans=0;    int num;        for (int i=0;i<9;i++) {num=0;        for (int j=i+1;j<9;j++) {if (ss[i]>ss[j]) num++;    } Ans+=num*f[i]; } return ans+1;}    void BFs () {int X,Y,T,T1,TEMPV;    Char temps[9];     memset (vis,0,sizeof Vis);    while (!q.empty ()) Q.pop ();    for (int i=0;i<8;i++) cur.a[i]=i+1-0+ ' 0 ';    cur.a[8]= ' 0 ';    cur.pos=9;    Cur.path= "";    Cur.val=kanto (CUR.A);    Vis[cur.val]=1;    Anspath[cur.val]=cur.path;    Q.push (cur); WHile (!q.empty ()) {Now=q.front ();        T=now.pos;        y=t%3;        if (y==0) y=3;        x= (t-y)/3+1;        Q.pop ();        int nx,ny;            for (int i=0;i<4;i++) {nx=x+dx[i];            ny=y+dy[i];//eyes are blind, just see written n=y=dy[i];(⊙o⊙) ... t1= (nx-1) *3+ny;            memcpy (temps,now.a,sizeof (temps)); TEMPS[T-1]=NOW.A[T1-1];            Analog mobile temps[t1-1]= ' 0 ';            Tempv=kanto (temps);                if (nx>=1 && nx<=3 && ny>=1 && ny<=3 &&!VIS[TEMPV]) {                Vis[tempv]=1;                CUR.POS=T1;                Cur.path=now.path+dir[i];                memcpy (cur.a,temps,sizeof (temps));                CUR.VAL=TEMPV;                Q.push (cur);            Anspath[cur.val]=cur.path;    }}}}int Main () {//f[0]=1;//for (int i=1;i<9;i++)//f[i]=f[i-1]*i;    BFS ();     while (cin>>c) {if (c== ' x ') {Cur.pos=1;       cur.a[0]= ' 0 ';        } else cur.a[0]=c;            for (int i=1;i<9;i++) {cin>>c;                if (c== ' x ') {cur.pos=i+1;            cur.a[i]= ' 0 ';        } else cur.a[i]=c;        } cur.val=kanto (CUR.A);        int flag;            if (Vis[cur.val]) {flag=anspath[cur.val].length ();        for (int i=flag-1;i>=0;i--) printf ("%c", Anspath[cur.val][i]);        } else printf ("unsolvable");    cout<<endl; } return 0;}

HUD 1043 Eight