Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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<cmath>
#include<string>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<map>
#include<queue>
using namespace std;
#define N 3
#define N2 9

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;
    }
};

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

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

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() {
    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;
    
    return 0;
}