Problem:
Score : 400 points

Problem Statement

Takahashi has a maze, which is a grid of H \times W squares with H horizontal rows and W vertical columns.

The square at the i-th row from the top and the j-th column is a "wall" square if S_{ij} is #, and a "road" square if S_{ij} is ..

From a road square, you can move to a horizontally or vertically adjacent road square.

You cannot move out of the maze, move to a wall square, or move diagonally.

Takahashi will choose a starting square and a goal square, which can be any road squares, and give the maze to Aoki.

Aoki will then travel from the starting square to the goal square, in the minimum number of moves required.

In this situation, find the maximum possible number of moves Aoki has to make.

Constraints

1 \leq H,W \leq 20

S_{ij} is . or #.

S contains at least two occurrences of ..

Any road square can be reached from any road square in zero or more moves.

Input

Input is given from Standard Input in the following format:

H W
S_{11}...S_{1W}
:
S_{H1}...S_{HW}

Output

Print the maximum possible number of moves Aoki has to make.

Sample Input 1

3 3
...
...
...

Sample Output 1

4

If Takahashi chooses the top-left square as the starting square and the bottom-right square as the goal square, Aoki has to make four moves.

Sample Input 2

3 5
...#.
.#.#.
.#...

Sample Output 2

10

If Takahashi chooses the bottom-left square as the starting square and the top-right square as the goal square, Aoki has to make ten moves.

Solution in Scala (2.11.7) (judge verdict: Wrong Answer)

import scala.collection.immutable.Queue

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val H, W = sc.nextInt(); sc.nextLine()
    val S = Array.fill(H)(sc.nextLine().trim.toCharArray)
    // スタートの決定にO(H*W)
    // ゴールの決定にO(H*W)
    // 最短路の探索にO(H*W)
    // 全探索するとO(H^3*W^3)
    // 64 * 10^6で間に合わない
    // dist(y)(x)(y')(x') := マスy,xからマスy',x'への最短路としておく
//    def solve(H: Int, W: Int, S: Array[Array[Char]]): Int = {
//      val MAX = Int.MaxValue
//      val dist = Array.fill(H, W, H, W)(MAX)
//      val dx = Array(0, 0, -1, 1)
//      val dy = Array(-1, 1, 0, 0)
//
//      def bfs(sx: Int, sy: Int, ex: Int, ey: Int): Int = {
//        val UNREACHED = 401
//        val arr = Array.fill(H, W)(MAX)
//        arr(sy)(sx) = 0
//        def loop(queue: Queue[(Int, Int)], min: Int): Int = {
//          if (queue.nonEmpty) {
//            val (x, y) = queue.head
//            if (x == ex && y == ey) {
//              val d = min min arr(ey)(ex)
//              dist(sy)(sx)(y)(x) = d
//              dist(y)(x)(sy)(sx) = d
//              d
//            } else if (dist(y)(x)(ey)(ex) != MAX) {
//              loop(queue.tail, min min (arr(y)(x) + dist(y)(x)(ey)(ex)))
//            } else {
//              if (min == UNREACHED) {
//                dist(sy)(sx)(y)(x) = arr(y)(x)
//                dist(y)(x)(sy)(sx) = arr(y)(x)
//              }
//              loop((0 to 3).foldLeft(queue.tail) { (queue, i) =>
//                val nx = x + dx(i)
//                val ny = y + dy(i)
//                if (nx >= 0 && nx < W &&
//                  ny >= 0 && ny < H &&
//                  S(ny)(nx) == '.' &&
//                  arr(ny)(nx) == MAX) {
//                  arr(ny)(nx) = arr(y)(x) + 1
//                  queue.enqueue((nx, ny))
//                } else queue
//              }, min)
//            }
//          } else min
//        }
//        loop(Queue((sx, sy)), UNREACHED)
//      }
//
//      (for {
//        sy <- 0 until H
//        sx <- 0 until W
//        if (S(sy)(sx) == '.')
//        ey <- (0 until H).reverse
//        ex <- (0 until W).reverse
//        if (S(ey)(ex) == '.')
//        if ey > sy || (ey == sy && ex > sx)
//      } yield bfs(sx, sy, ex, ey)).max
//    }

    // 開始地点と終了地点を全探索していたが, そもそも開始地点を確定させて
    // BFSをすると, 一回探索した時に全ての終了地点までの最短経路が求まるので
    // 開始地点の列挙O(HW)に、BFSで最後まで探索した時の距離 O(HW）を求め、その最大を求めればいい
    def solve(H: Int, W: Int, S: Array[Array[Char]]): Int = {
      val dx = Array(0, 0, -1, 1)
      val dy = Array(1, -1, 0, 0)
      def bfs(sx: Int, sy: Int): Int = {
        val visited = Array.fill(H, W)(false)
        def loop(queue: Queue[((Int, Int), Int)], max: Int): Int =
          if (queue.nonEmpty) {
            val ((x, y), dist) = queue.head
            loop((0 until 4).foldLeft(queue.tail)((queue, i) => {
              val nx = x + dx(i)
              val ny = y + dy(i)
              if (nx >= 0 && nx < W &&
                  ny >= 0 && ny < H &&
                  S(ny)(nx) == '.' &&
                  !visited(ny)(nx)) {
                visited(ny)(nx) = true
                queue.enqueue(((nx, ny), dist + 1))
              } else queue
            }), max max dist)
          } else max
        loop(Queue(((sx, sy), 0)), 0)
      }

      (for {
        sy <- 0 until H
        sx <- 0 until W
        if S(sy)(sx) == '.'
      } yield bfs(sx, sy)).max
    }

    println(solve(H, W, S))
  }

}