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++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

struct Puzzle{
    int f[9];
    int space;
    string path;

    bool operator<(const Puzzle &p)const{
        for(int i=0;i<9;i++){
            if(f[i]==p.f[i]) continue;
            return f[i]<p.f[i];
        }
        return false;
    }
};

bool isTarget(Puzzle &p){
    for(int i=0;i<9;i++){
        if(p.f[i]!=i+1) return false;
    }
    return true;
}

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'};

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/3;
        int sy=u.space%3;
        for(int r=0;r<4;r++){
            int tx=sx+dx[r];
            int ty=sy+dy[r];
            if(tx<0||ty<0||tx>2||ty>2){
                continue;
            }
            v=u;
            swap(v.f[u.space],v.f[tx*3+ty]);
            v.space=tx*3+ty;
            if(!V[v]){
                V[v]=true;
                v.path+=dir[r];
                Q.push(v);
            }
        }
    }
    return "unsolvable";
}

int main(){
    Puzzle x;
    for(int i=0;i<9;i++){
        cin>>x.f[i];
        if(x.f[i]==0){
            x.f[i]=9;
            x.space=i;
        }
    }

    cout<<(bfs(x)).size()<<endl;
}