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 <cstdio>
#include <climits>
#include <cmath>
#include <cassert>
#include <algorithm>
#include <utility>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <map>

#define MAX 10000
#define INFTY 1<<30 // 2^30
#define N 3
#define N2 9  // N * N
#define REP(i,n) for(int i=0;i<n;i++)

/*
priority_queue<> pq
 push()  pop() size()  top()
map<>
 find() begin() end() insert(map<>::value_type())
pair<>
sort(.begin(), .end())
sort(arr, arr+N)
sort(a, a+N, greater<int>())
binary_search(vc.begin(), vc.end(), x)  //binary_search???true, false?????????
vector<int>::iterator lit = lower_bound(vc.begin(), vc.end(), x);
  lower_bound?????????????????????"??\???"????????????????????????????????????????????¬???????????????
vector<int>::iterator uit = upper_bound(vc.begin(), vc.end(), x);
  upper_bound????????????????????????"?¶???????"?????????????????????????????????????????¬???????????????
*/


using namespace std;

typedef long long llong;

struct Puzzle
{
  int f[N2];
  int space;
  string path;

  bool operator < (const Puzzle &p) const
  {
    for(int i = 0; i < N2; i++)
    {
      if(f[i] == p.f[i]) continue;
      return f[i] > p.f[i];
    }
    return false;
  }
};
/*
  1 3 2       1 2 3
  4 5 6   <   4 5 6
  7 8 9       7 8 9
  : true
  1 2 3       1 3 2
  4 5 6   <   4 5 6
  7 8 9       7 8 9
  : false
  ??¨??¨??????
  : false
*/

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

bool isTarget(Puzzle p)
{
  for(int i = 0; i< N2; i++)
    if(p.f[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.space / N;
    int sy = u.space % 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.f[u.space], v.f[tx*N+ty]);
      v.space = tx*N+ty;
      if(!V[v])
      {
        V[v] = true;
        v.path += dir[r];
        Q.push(v);
      }
    }
  }
  return "unsolvable";
}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  // FROM HERE
  Puzzle in;
  for(int i = 0; i < N2; i++)
  {
    cin >> in.f[i];
    if(in.f[i] == 0)
    {
      in.f[i] = N2;
      in.space = i;
    }
  }
  string ans = bfs(in);
  cout << ans.size() << endl; 
  //cerr << bfs(in) << endl;
  // END
  return 0;
}