Problem:
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15 Puzzle

The goal of the 15 puzzle problem is to complete pieces on $4 \times 4$ cells where one of the cells is empty space.

In this problem, the space is represented by 0 and pieces are represented by integers from 1 to 15 as shown below.

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

You can move a piece toward the empty space at one step. Your goal is to make the pieces the following configuration in the shortest move (fewest steps).

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 0

Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Input

The $4 \times 4$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

The given puzzle is solvable in at most 45 steps.

Sample Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

Sample Output

8

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 4
#define N2 16
#define LIMIT 100

static const int dy[4] = { 1, 0, -1, 0};
static const int dx[4] = { 0, -1, 0, 1};
static const char dir[4] = { 'r', 'u', 'l', 'd'};
int MDT[N2][N2];

typedef struct p{
  int f[N2];
  int space;
  int MD;
} Puzzle;

Puzzle state;
int lim;
int path[LIMIT];

int getAllMD(Puzzle p){
  int sum=0, i;
  for( i = 0 ; i < N2 ; i++ ){
    if( p.f[i] == N2 ) continue;
    sum += MDT[i][p.f[i]-1];
  }
  return sum;
}

int isSolved(){
  int i;
  for( i = 0 ; i < N2 ; i++ ){
    if( state.f[i] != i+1 ) return 0;
  }
  return 1;
}

int dfs(int depth, int prev){
  int r;
  Puzzle tmp;
  if( state.MD == 0 ) return 1;
  if( depth + state.MD > lim ) return 0;

  int sx = state.space / N;
  int sy = state.space % N;

  for( r = 0 ; r < 4 ; r++ ){
    int tx = sx + dx[r];
    int ty = sy + dy[r];
    if( tx < 0 || ty < 0 || tx >= N || ty >= N ) continue;
    if( (fmax( prev, r ) - fmin( prev, r )) == 2 ) continue;
    tmp = state;
    state.MD -= MDT[tx * N + ty][state.f[tx * N + ty]-1];
    state.MD += MDT[sx * N + sy][state.f[tx * N + ty]-1];
    int t;
    t = state.f[tx * N + ty];
    state.f[tx * N + ty] = state.f[sx * N + sy];
    state.f[sx * N + sy] = t;
    state.space = tx * N + ty;
    if( dfs(depth+1,r) == 1 ){
      path[depth] = r;
      return 1;
    }
    state = tmp;
  }
  return 0;
}

char ans[1000];

int iterative_deepening(Puzzle in){
  int i;

  in.MD = getAllMD(in);

  for( lim = in.MD ; lim <= LIMIT ; lim++ ){
    state = in;
    if( dfs(0, -100) == 1 ){
      ans[0] = '\0';
      for( i = 0 ; i < lim ; i++ ) ans[i+1] = dir[path[i]];
      return i;
    }
  }
  ans[10] = '\0';
  return 10;
}

int main(){
  int i, j, ans;

  for( i = 0 ; i < N2 ; i++ ){
    for( j = 0 ; j < N2 ; j++ ){
      MDT[i][j] = fabs( i / N - j / N ) + fabs( i % N - j % N );
    }
  }
  Puzzle in;
  for( i = 0 ; i < N2 ; i++ ){
    scanf("%d",&in.f[i]);
    if( in.f[i] == 0 ){
      in.f[i] = N2;
      in.space = i;
    }
  }
  ans = iterative_deepening(in);
  printf("%d\n",ans);

  return 0;
}