uva 10422 - Knights in FEN

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 expect?).
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
First line of the input file contains an integer N (N<14) that indicates how many sets of inputs are there. The description of each set is given below:
Each set consists of five lines; each line represents one row of a chessboard. The positions occupied by white knights are marked by 0 and the positions occupied by black knights are marked by 1. The space corresponds to the empty square on board.
There is no blank line between the two sets of input.
The first set of the sample input below corresponds to this configuration:
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 in n move(s).
where n <= 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.”
정통적인 방법은 bfs+hash 판정 중복이지만 제목은 최대 걸음 수 제한을 주고 dfs+로 가지를 빼서
최대 4^11, 5*5는 매번 4개 방향을 확장하지 않기 때문에 데이터 규모가 크지 않기 때문에 중복 판정을 할 필요가 없습니다. 우리는 매번 되돌아가지 않으면 됩니다. 이렇게 하면 중복 상황이 존재하지만 불필요한 검색을 많이 줄일 수 있습니다. 0.008sAC

#include<stdio.h>
#include<string.h>
int a[5][5],move[8][2]={{-1,-2},{-1,2},{-2,1},{-2,-1},{2,1},{2,-1},{1,-2},{1,2}},
    b[5][5]={{1,1,1,1,1},{0,1,1,1,1},{0,0,2,1,1},{0,0,0,0,1},{0,0,0,0,0}},min;
int dfs(int x,int y,int step,int way)
{int i,j,X,Y,sum=0,t;
 if (step>=min) return 0;
 for (i=0;i<5;i++)
 for (j=0;j<5;j++)
 if (a[i][j]!=b[i][j]) ++sum;
 if (sum==0) {if (step<min) min=step; return 0;}
 if ((sum+1)/2+step>=min) return 0; //SUM ， ( 2 ）sum/2+sum%2=（sum+1）/2 
 for (i=0;i<8;i++)
 {X=x+move[i][0];
  Y=y+move[i][1];
  if ((X>=0)&&(X<5)&&(Y>=0)&&(Y<5)&&(way+i!=7)) // move 7， 
  {t=a[x][y];a[x][y]=a[X][Y];a[X][Y]=t;
   dfs(X,Y,step+1,i);
   t=a[x][y];a[x][y]=a[X][Y];a[X][Y]=t;
  }
 }
}
int main()
{int i,j,x,y,n;
 char s[6];
 scanf("%d
",&n);
 while (n--)
 {
  for (i=0;i<5;i++)
  {gets(s);
   for (j=0;s[j]!='\0';j++)
   if (s[j]==' ') {a[i][j]=2;x=i;y=j;}
             else a[i][j]=s[j]-'0';
  }
  min=11;
  dfs(x,y,0,8);
  if (min==11) printf("Unsolvable in less than 11 move(s).
");
  else printf("Solvable in %d move(s).
",min);
 }
 return 0;
}