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

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

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

puzzle state;
int limit;
int path[LIMIT];

void swap(int *a, int *b)
{
    int tmp;
    tmp = *a;
    *a = *b;
    *b = tmp;
}
int getALLMD(puzzle pz)
{
    int sum = 0;
    for (int i = 0; i < N2; i++)
    {
        if (pz.f[i] == N2)
            continue;
        sum += MDT[i][pz.f[i] - 1];
    }
    return sum;
}

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

bool dfs(int depth, int prev)
{
    if (!state.MD)
        return true;
    if (depth + state.MD > limit)
        return false;

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

    for (int 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];
        swap(&state.f[tx * N + ty], &state.f[sx * N + sy]);
        state.space = tx * N + ty;
        if (dfs(depth + 1, r))
        {
            path[depth] = r;
            return true;
        }
        state = tmp;
    }
    return false;
}

int deepening(puzzle in)
{
    in.MD = getALLMD(in);

    for (limit = in.MD; limit <= LIMIT; limit++)
    {
        state = in;
        if (dfs(0, -100))
        {
            char a[LIMIT] = {};
            for (int i = 0; i < limit; i++)
                a[i] += dir[path[i]];
            return strlen(a);
        }
    }

    return false;
}

int main()
{
    for (int i = 0; i < N2; i++)
        for (int j = 0; j < N2; j++)
            MDT[i][j] = fabs(i / N - j / N) + fabs(i % N - j % N);

    puzzle in;
    for (int i = 0; i < N2; i++)
    {
        scanf("%d", &in.f[i]);
        if (!in.f[i])
        {
            in.f[i] = N2;
            in.space = i;
        }
    }
    printf("%d\n", deepening(in));
    if (!isSolved())
        printf("failed\n");
    return 0;
}