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 Python3 (judge verdict: Time Limit Exceeded)

import sys;
import heapq

def iterative(i,j):
    q = []
    heapq.heappush(q,(0,i,j,0,puz,costs))
    global finding

    while len(q):
        items = heapq.heappop(q)
        c_depth = items[0]
        _i = items[1]
        _j = items[2]
        prev_move = items[3]
        c_puz = items[4]
        _costs = items[5]
        if(puzzleCheck(c_puz)):
            finding = 1
            print(c_depth)
            break

        if((c_depth+sumCost(_costs)) >= depth):
            continue

        if(_i != 0 and prev_move != 1):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i-1,_j)
            swap_cost = _costs[0:]
            swap_cost[_i*3+_j] = HS(_i,_j,swap_puz[_i*3+_j])
            swap_cost[(_i-1)*3+_j] = HS(_i-1,_j,swap_puz[(_i-1)*3+_j])
            if(not checkPuzIsExisted(swap_puz,q)):
                heapq.heappush(q,(c_depth+1,_i-1,_j,2,swap_puz,swap_cost))
        if(_i != 2 and prev_move != 2):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i+1,_j)
            swap_cost = _costs[0:]
            swap_cost[_i*3+_j] = HS(_i,_j,swap_puz[_i*3+_j])
            swap_cost[(_i+1)*3+_j] = HS(_i+1,_j,swap_puz[(_i+1)*3+_j])
            if(not checkPuzIsExisted(swap_puz,q)):
                heapq.heappush(q,(c_depth+1,_i+1,_j,1,swap_puz,swap_cost))
        if(_j != 0 and prev_move != 3):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i,_j-1)
            swap_cost = _costs[0:]
            swap_cost[_i*3+_j] = HS(_i,_j,swap_puz[_i*3+_j])
            swap_cost[_i*3+_j-1] = HS(_i,_j-1,swap_puz[_i*3+_j-1])
            if(not checkPuzIsExisted(swap_puz,q)):
                heapq.heappush(q,(c_depth+1,_i,_j-1,4,swap_puz,swap_cost))
        if(_j != 2 and prev_move != 4):
            swap_puz = swapPuz(c_puz[0:],_i,_j,_i,_j+1)
            swap_cost = _costs[0:]
            swap_cost[_i*3+_j] = HS(_i,_j,swap_puz[_i*3+_j])
            swap_cost[_i*3+_j+1] = HS(_i,_j+1,swap_puz[_i*3+_j+1])
            if(not checkPuzIsExisted(swap_puz,q)):
                heapq.heappush(q,(c_depth+1,_i,_j+1,3,swap_puz,swap_cost))

def checkPuzIsExisted(puz,q):
    for i in range(len(q)):
        exist_puz = q[i][4]
        count = 0
        for j in range(len(exist_puz)):
            if(exist_puz[j] != puz[j]):
                count = 1
        if(count == 0):
            return True
    return False

def sumCost(costs):
    value = 0
    for i in range(len(costs)):
        value += costs[i]
    return value

def simpleHS(c_puz):
    count = 0
    for i in range(len(correctPuz)):
        if(correctPuz[i] != c_puz[i]):
            count+=1
    return count

def HS(i,j,num):
    k = correctPuz.index(num)
    ki = (int)(k/3)
    kj = k - ki*3
    value = abs(_i-ki)+abs(_j-kj) 
    return value


def swapPuz(c_puz, i, j, i2,j2):
    c_puz[i2*3+j2],c_puz[i*3+j] = c_puz[i*3+j],c_puz[i2*3+j2]
    return c_puz

def puzzleCheck(c_puz):
    if(c_puz[8] == 0):
        for i in range(len(correctPuz)):
            if(correctPuz[i] != c_puz[i]):
                return False
        return True
    return False

correctPuz = [i+1 for i in range(9)]
correctPuz[8] = 0
puz = [0 for i in range(9)]
costs = [0 for i in range(9)]

i_start = 0
j_start = 0

for i in range(3):
    puz[i*3],puz[i*3+1],puz[i*3+2] = map(int, input().split());
    if(puz[i*3] == 0):
        i_start,j_start = i,0
    elif(puz[i*3+1] == 0):
        i_start,j_start = i,1
    elif(puz[i*3+2] == 0):
        i_start,j_start = i,2

for i in range(len(costs)):
    _i = (int)(i/3)
    _j = i - _i*3
    costs[i] = HS(_i,_j,puz[i])

finding = 0
depth = 0
while True:
    if(finding == 1):
        break
    iterative(i_start,j_start)
    depth+=1