hdu.1043.Eight (打表 || 双广 + 奇偶逆序)

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14380    Accepted Submission(s): 4044 Special Judge

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 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, 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

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<ctype.h>
 typedef long long ll ;
 int map[][] ;
 char mp[] ;
 int l , r , l1 , r1 ;
 int flag ;
 int anti ;
 int move[][] = {{,} , {-,} , {,} , {,-}} ;
 const int mod =  ;
 struct node
 {
     int x , y , nxt ;
     int map[][] ;
 } b[mod];
 struct hash
 {
     ll w , id ;
     int nxt ;
 }e[mod * ];
 int H[mod * ] , E ;
 
 void insert (ll x , int id)
 {
     int y  = x % mod ;
     if (y < ) y += mod ;
     e[++ E].w = x ;
     e[E].id = id ;
     e[E].nxt = H[y] ;
     H[y] = E ;
 }
 
 int find (ll x)
 {
     int y = x % mod ;
     if (y <  ) y += mod ;
     for (int i = H[y] ; ~ i ; i = e[i].nxt ) {
         if (e[i].w == x)
             return e[i].id ;
     }
     return - ;
 }
 
 void table ()
 {
     node ans , tmp ;
     E = - ;  memset (H , - , sizeof (H) ) ;
     l1 =  , r1=  ;
     b[l1].x =  , b[l1].y =  , b[l1].nxt = - ;
     ll sum =  ;
     for (int i =  ; i <  ; i ++) for (int j =  ; j <  ; j ++) b[l1].map[i][j] = i *  + j +  ;
     insert ( ,  ) ;
     while (l1 != r1) {
         ans = b[l1] ;
         for (int i =  ; i <  ; i ++) {
             tmp.x = ans.x + move[i][] ; tmp.y = ans.y + move[i][] ;
             if (tmp.x <  || tmp.y <  || tmp.x >=  || tmp.y >= ) continue ;
             for (int i =  ; i <  ; i ++) for (int j =  ; j <  ; j ++) tmp.map[i][j] = ans.map[i][j] ;
             std::swap (tmp.map[ans.x][ans.y] , tmp.map[tmp.x][tmp.y]) ;
             sum =  ;
             for (int i =  ; i <  ; i ++) for (int j =  ; j <  ; j ++) sum = sum *  + tmp.map[i][j] ;
             if (find (sum) != - ) continue ;
             insert (sum  , r1 ) ;
             tmp.nxt = l1 ;
             b[r1 ++] = tmp ;
         }
         l1 ++ ;
     }
 }
 
 void solve (int u)
 {
     if (u == -) return ;
     int t = b[u].nxt ;
     if (t != -) {
         int x = b[t].x - b[u].x , y = b[t].y - b[u].y ;
         if (x == ) printf ("d") ;
         else if (x == -) printf ("u") ;
         else if (y == ) printf ("r") ;
         else if (y == - ) printf ("l") ;
     }
     solve (t) ;
 }
 
 int main ()
 {
     freopen ("a.txt" , "r" , stdin ) ;
     table () ;
     while (gets (mp) != NULL) {
         int tot =  ;
         for (int i =  ; mp[i] != '\0' ; i ++) {
             if (mp[i] != ' ') {
                 if (isdigit (mp[i])) {
                     map[tot / ][tot % ] = mp[i] - '' ;
                 }
                 else map[tot / ][tot % ] =  ;
                 tot ++ ;
             }
         }
         ll sum =  ;
         for (int i =  ; i <  ; i ++) for (int j =  ; j <  ; j ++) sum = sum *  + map[i][j] ;
         anti = find (sum) ;
         if (anti != -) solve (anti) ;
         else printf ("unsolvable") ;
         puts ("") ;
     }
     return  ;
 }

打表思路很简单,也是为了对抗unsolve这种情况,从123456789最终状态产生所有状态,即可.

奇偶逆序:

逆序数:也就是说，对于n个不同的元素，先规定各元素之间有一个标准次序（例如n个 不同的自然数，可规定从小到大为标准次序），于是在这n个元素的任一排列中，当某两个元素的先后次序与标准次序不同时，就说有1个逆序。一个排列中所有逆序总数叫做这个排列的逆序数。

结论:(是除去移动元素,(这里是9))

对源状态A与目标状态B进行规范化，使得两矩阵的元素0的位置相同；记为新的源状态A'与目标状态B';
若A'与B'的逆序对的奇偶性相同（即A'与B1的逆序对的奇偶性相同），则A'必定可能转化为B',即A可以转化到B; 
若A'与B'的逆序对的奇偶性不同（即A'与B2的逆序对的奇偶性相同），则A'必定不可能转化为B',即A不可以转化到B;

也就是为了对抗unsolve.

然后用双广就OK了.