Knights in FEN UVA10422

Problem D

Knights in FEN

Input:Standard input

Output:Standard output

Time Limit:10 seconds

There are black and white knights on a 5 by 5 chessboard. there are twelve of each color, and there is one square that is empty. at any time, a knight can move into an empty square as long as it moves like a knight in normal chess (what else did you have CT ?).

Given an initial position of the board, the question is: what is the minimum number of moves in which we can reach the final position which is:

Input <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> Vc3ryb25np1_vcd4kpha + ratio = "206" height = "207" src = "http://www.bkjia.com/uploadfile/Collfiles/20140107/20140107095121228.jpg" alt = "\">

Output

For each set your task is to find the minimum number of moves leading from the starting input configuration to the final one. If that number is bigger than 10, then output one line stating

Unsolvable in less than 11 move(s).

Otherwise output one line stating

Solvable inNMove (s ).

WhereN<= 10.

The output for each set is produced in a single line as shown in the sample output.

Sample Input

2

01011

110 1

01110

01010

00100

10110

01 11

10111

01001

00000

Sample Output

Unsolvable in less than 11 move(s).

Solvable in 7 move(s).

(Problem Setter: Piotr Rudnicki, University of Alberta, Canada)

"A man is as great as his dreams ."

A status chart is provided to move the server guard to the initial state.

Typical Implicit Graph Search problems: BFS search + hash. Hash is implemented using set. In addition, it should be noted that the server guard in chess uses Japanese characters.

# Include
  
   
# Include
   
    
# Include
    
     
# Include
     
      
Using namespace std; typedef int state [25]; const int maxn = 5000000; const int dx [] = {,-1,-2,-2, -1}; const int dy [] = {-2,-, 1,-1,-2}; state st [maxn]; int dist [maxn]; int front, rear, s; set
      
       
Vis; state goal = {, 0, 0}; int try_to_insert (int s) // hash function {int v = 0; for (int I = 0; I <25; I ++) v = v * 2 + st [s] [I]; if (vis. count (v) return 0; vis. insert (v); return 1;} int bfs () {front = 1, rear = 2; vis. clear (); while (front
       
         10) return-1; // pruning. if (memcmp (goal, p, sizeof (p) = 0) return front; int I, j, x, y; for (I = 0; I <25; I ++) if (st [front] [I] = 2) break; int z = I; x = I/5, y = I % 5; for (I = 0; I <8; I ++) {int newx = x + dx [I]; int newy = y + dy [I]; if (newx> = 0 & newx <5 & newy> = 0 & newy <5) {state & u = st [rear]; memcpy (& u, & p, sizeof (p )); u [x * 5 + y] = u [newx * 5 + newy]; u [newx * 5 + newy] = 2; dist [rear] = dist [front] + 1; if (try_to_insert (rear) rear ++;} fr Ont ++ }}int main () {cin> s; string str; getline (cin, str); while (s --) {memset (dist, 0, sizeof (dist); int I, j; for (I = 0; I <5; I ++) {getline (cin, str); for (j = 0; j <5; j ++) {if (str [j]! = '') St [1] [I * 5 + j] = str [j]-'0 '; else st [1] [I * 5 + j] = 2 ;}} int d = bfs (); if (d <= 0) cout <"Unsolvable in less than 11 move (s ). "<