Problem:
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8 Puzzle

The goal of the 8 puzzle problem is to complete pieces on $3 \times 3$ 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 8 as shown below.

1 3 0
4 2 5
7 8 6

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 0

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

Input

The $3 \times 3$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

There is a solution.

Sample Input

1 3 0
4 2 5
7 8 6

Sample Output

4

Solution in C++ (judge verdict: Accepted)

#include<iostream>
#include<map>
#include<queue>
#include<string>
using namespace std;

static const int N = 3;
static const int N2 = 9;

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

struct Puzzle {
  int pas[N2], ss;
  string path;

  bool operator < (const Puzzle &p1) const {

    for (int i = 0; i < N2; i++) {
      if (pas[i] == p1.pas[i]) continue;
      return pas[i] > p1.pas[i];
    }
    return false;
  }
};

bool isTarget(Puzzle u) {
  for (int i = 0; i < N2; i++) {
    if (u.pas[i] != i + 1) return false;
  }
  return true;
}

string bfs(Puzzle s) {
  queue<Puzzle> Q;
  map <Puzzle, bool> V;

  Puzzle u, v;
  s.path = "";
  Q.push(s);
  V[s] = true;
  while (!Q.empty()) {
    u = Q.front();
    Q.pop();
    if (isTarget(u)) return u.path;
    int sx = u.ss / N;
    int sy = u.ss % N;
    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;
      v = u;
      swap(v.pas[u.ss], v.pas[tx * N + ty]);
      v.ss = tx * N + ty;
      if (!V[v]) {
	V[v] = true;
	v.path += dir[r];
	Q.push(v);
      }
    }
  }
  return "unsolvable";
}


int main() {
  Puzzle in;
  for (int i = 0; i < N2; i++) {
    cin >> in.pas[i];
    if (!in.pas[i]) {
      in.pas[i] = N2;
      in.ss = i;
    }
  }

  string ans = bfs(in);
  cout << ans.size() << endl;

  return 0;
}