Problem:
Problem C: Push!!

Mr. Schwarz was a famous powerful pro wrestler. He starts a part time job as a warehouseman.
His task is to move a cargo to a goal by repeatedly pushing the cargo in the warehouse, of course,
without breaking the walls and the pillars of the warehouse.

There may be some pillars in the warehouse. Except for the locations of the pillars, the floor of
the warehouse is paved with square tiles whose size fits with the cargo. Each pillar occupies the
same area as a tile.

Initially, the cargo is on the center of a tile. With one push, he can move the cargo onto the
center of an adjacent tile if he is in proper position. The tile onto which he will move the cargo
must be one of (at most) four tiles (i.e., east, west, north or south) adjacent to the tile where
the cargo is present.

To push, he must also be on the tile adjacent to the present tile. He can only push the cargo in
the same direction as he faces to it and he cannot pull it. So, when the cargo is on the tile next
to a wall (or a pillar), he can only move it along the wall (or the pillar). Furthermore, once he
places it on a corner tile, he cannot move it anymore.

He can change his position, if there is a path to the position without obstacles (such as the cargo
and pillars) in the way. The goal is not an obstacle. In addition, he can move only in the four
directions (i.e., east, west, north or south) and change his direction only at the center of a tile.

As he is not so young, he wants to save his energy by keeping the number of required pushes
as small as possible. But he does not mind the count of his pedometer, because walking is very
light exercise for him.

Your job is to write a program that outputs the minimum number of pushes required to move
the cargo to the goal, if ever possible.

Input

The input consists of multiple maps, each representing the size and the arrangement of the
warehouse. A map is given in the following format.

w h
d11 d12 d13 ... d1w
d21 d22 d23 ... d2w
...
dh1 dh2 dh3 ... dhw

The integers w and h are the lengths of the two sides of the floor of the warehouse in terms of
widths of floor tiles. w and h are less than or equal to 7. The integer dij represents what is
initially on the corresponding floor area in the following way.

0: nothing (simply a floor tile)

1: a pillar

2: the cargo

3: the goal

4: the warehouseman (Mr. Schwarz)

Each of the integers 2, 3 and 4 appears exactly once as dij in the map. Integer numbers in an
input line are separated by at least one space character. The end of the input is indicated by a
line containing two zeros.

Output

For each map, your program should output a line containing the minimum number of pushes.
If the cargo cannot be moved to the goal, -1 should be output instead.

Sample Input

5 5
0 0 0 0 0
4 2 0 1 1
0 1 0 0 0
1 0 0 0 3
1 0 0 0 0
5 3
4 0 0 0 0
2 0 0 0 0
0 0 0 0 3
7 5
1 1 4 1 0 0 0
1 1 2 1 0 0 0
3 0 0 0 0 0 0
0 1 0 1 0 0 0
0 0 0 1 0 0 0
6 6
0 0 0 0 0 3
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 2 0 0 0 0
4 0 0 0 0 0
0 0

Output for the Sample Input

5
-1
11
8

Solution in JAVA (judge verdict: Accepted)

import java.util.*;
import java.awt.Point;

public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		while(true){
			w = sc.nextInt();
			h = sc.nextInt();
			if(w == 0 && h == 0) break;

			t = new int[h][w];
			Point p = null;
			Point b = null;

			for(int i=0;i<h;i++){
				for(int j=0;j<w;j++){
					t[i][j] = sc.nextInt();
					if(t[i][j] == 4){
						p = new Point(j,i);
						t[i][j] = 0;
					}
					else if(t[i][j] == 2){
						b = new Point(j,i);
						t[i][j] = 0;
					}
				}
			}

			PriorityQueue<State> open = new PriorityQueue<State>();
			HashSet<State> closed = new HashSet<State>();
			State ans = null;
			open.add(new State(p,b,0));

			while(!open.isEmpty()){
				State st = open.poll();
				if(closed.contains(st)) continue;
				closed.add(st);

				if(t[st.b.y][st.b.x] == 3){
					ans = st;
					break;
				}

				open.addAll(st.nexts());
			}

			System.out.println(ans == null ? -1 : ans.cost);
		}
	}

	private static int w,h;
	private static int[][] t;
	private static int[] dx = {0,1,0,-1};
	private static int[] dy = {-1,0,1,0};

	private static boolean reachable(Point s,Point g,Point b){
		t[b.y][b.x] = 2;

		Queue<Point> open = new LinkedList<Point>();
		HashSet<Point> closed = new HashSet<Point>();
		open.add(s);

		while(!open.isEmpty()){
			Point p = open.poll();
			if(closed.contains(p)) continue;
			closed.add(p);
			if(p.equals(g)){
				t[b.y][b.x] = 0;
				return true;
			}

			for(int i=0;i<4;i++){
				int nx = p.x + dx[i];
				int ny = p.y + dy[i];

				if(nx>=0 && nx<w && ny>=0 && ny<h && (t[ny][nx]==0 || t[ny][nx]==3)){
					open.add(new Point(nx,ny));
				}
			}
		}

		t[b.y][b.x] = 0;

		return false;
	}

	private static class State implements Comparable<State>{
		Point p,b;
		int cost;

		State(Point p,Point b,int cost){
			this.p = new Point(p.x,p.y);
			this.b = new Point(b.x,b.y);
			this.cost = cost;
		}

		List<State> nexts(){
			List<State> sts = new ArrayList<State>();

			for(int i=0;i<4;i++){
				int nx1 = b.x + dx[i];
				int ny1 = b.y + dy[i];
				int nx2 = b.x + dx[(i+2)%4];
				int ny2 = b.y + dy[(i+2)%4];
				Point np1 = new Point(nx1,ny1);
				Point np2 = new Point(nx2,ny2);

				if(nx1>=0 && nx1<w && ny1>=0 && ny1<h &&
					 nx2>=0 && nx2<w && ny2>=0 && ny2<h &&
					 t[ny1][nx1]!=1 && t[ny2][nx2]!=1 && reachable(p,np1,b)){
					sts.add(new State(b,np2,cost+1));
				}
			}

			return sts;
		}

		public boolean equals(Object o){
			State st = (State)o;
			return p.equals(st.p) && b.equals(st.b);
		}

		public int hashCode(){
			return p.x + p.y * 10 + b.x * 100 + b.y * 1000;
		}

		public int compareTo(State st){
			return this.cost - st.cost;
		}
	}
}