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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class Main {

    public static void main(String[] args) throws IOException {
        new Main().solve();
    }

    private void solve() throws IOException {
        try {
//            solveA();
//            solveB();
//            solveC();
            solveD();
            //			 solveE();
            //			 solveF();
        } finally {
            if (in != null) {
                in.close();
            }
            if (out != null) {
                out.flush();
                out.close();
            }
        }

    }

    private void solveA() {
        char[] c = next().toCharArray();

        out.println((char) (c[0] + 1));
    }

    private void solveB() {
        int n = nextInt();
        int k = nextInt();
        int m = nextInt();
        int[] wk = IntStream.range(0, n - 1).map(i -> nextInt()).toArray();
        int diff = n * m - Arrays.stream(wk).sum();
        if (diff > k)
            out.println(-1);
        else
            out.println(diff > 0 ? diff : 0);

    }

    private static class Problem {
        int ind;
        int problem;
        int ans;

        public Problem(int ind, int problem, int ans) {
            this.ind = ind;
            this.problem = problem;
            this.ans = ans;
        }

    }

    private void solveC() {
        int n = nextInt();
        int m = nextInt();
        Problem[] problem = new Problem[m];
        for (int i = 0; i < m; i++) {
            problem[i] = new Problem(i, nextInt(), next().equals("WA") ? 0 : 1);
        }
        if (problem.length == 0) {
            out.println(0 + " " + 0);
            return;
        }
        Arrays.sort(problem, (x, y) -> Integer.compare(x.problem, y.problem) == 0 ? Integer.compare(x.ind, y.ind) : Integer.compare(x.problem, y.problem));
        Set<Integer> ac = new HashSet<Integer>();
        boolean notAc = true;
        int prev = problem[0].problem;
        Map<Integer, Integer> waMemo = new HashMap<Integer, Integer>();
        for (Problem p : problem) {
            if (prev != p.problem) {
                prev = p.problem;
                notAc = true;
            }
            if (notAc && p.ans == 0) {
                waMemo.put(p.problem, waMemo.getOrDefault(p.problem, 0) + 1);
            } else if (p.ans == 1) {
                notAc = false;
                ac.add(p.problem);
            }
        }

        int wa = 0;
        for (int a : ac) {
            wa += waMemo.getOrDefault(a, 0);
        }
        out.println(ac.size() + " " + wa);
    }

    private static class Field {
        int h;
        int w;
        int num;

        public Field(int h, int w, int num) {
            this.h = h;
            this.w = w;
            this.num = num;
        }
    }

    private void solveD() {
        int h = nextInt();
        int w = nextInt();
        char[][] map = Stream.generate(() -> next().toCharArray()).limit(h).toArray(char[][]::new);

        int max = 0;
        int[] mH = new int[]{1, -1, 0, 0};
        int[] mW = new int[]{0, 0, 1, -1};
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                if (map[i][j] == '#')
                    continue;
                boolean[][] visited = new boolean[h][w];
                visited[i][j] = true;
                Field start = new Field(i, j, 0);
                Queue<Field> queue = new ArrayDeque<Field>();
                queue.add(start);
                while (queue.size() > 0) {
                    int count = queue.size();
                    for (int k = 0; k < count; k++) {
                        Field wk = queue.poll();
                        int wkCount = wk.num + 1;
                        for (int l = 0; l < 4; l++) {
                            int wkH = mH[l] + wk.h;
                            int wkW = mW[l] + wk.w;

                            if (wkH < 0 || wkH >= h || wkW < 0 || wkW >= w)
                                continue;
                            if (visited[wkH][wkW])
                                continue;
                            visited[wkH][wkW] = true;

                            if (map[wkH][wkW] == '#')
                                continue;
                            queue.add(new Field(wkH, wkW, wkCount));
                            max = Math.max(max, wkCount);
                        }
                    }
                }
            }
        }

        out.println(max);
    }

    private void solveE() {
        int n = nextInt();

        out.println("");
    }

    private void solveF() {
        int n = nextInt();

        out.println("");
    }

    /*
     * 階乗計算の最大値
     * もっと大きいのが必要ならここを修正
     */
    private final int FACTORIAL_NUM = 111111;
    /*
     * 階乗の事前計算（mod 済）
     */
    private long[] factorial = new long[FACTORIAL_NUM];
    /*
     * 逆数の事前計算（mod 済）
     */
    private long[] reverseFactorial = new long[FACTORIAL_NUM];
    private boolean isInitFactorial = false;

    /**
     * 階乗計算の事前累積和
     * [1, 1, 2, 3, 4, 5, … FACTORIAL_NUM]
     * mod済
     */
    private void initFactorial(long mod) {
        factorial[0] = 1;
        for (int i = 1; i < FACTORIAL_NUM; i++) {
            factorial[i] = (factorial[i - 1] * i) % mod;
        }
    }

    /**
     * 逆数の事前計算
     * もっと速い方法があったので修正
     */
    private void initReverseFactorial(long mod) {

        reverseFactorial[FACTORIAL_NUM - 1] = modInverse(factorial[FACTORIAL_NUM - 1], mod);
        for (int i = FACTORIAL_NUM - 2; i >= 0; i++) {
            /*
             * 以下を利用する。
             * これにより毎回modInverseしなくてよくなる
             * (N!)^(-1) * N -> ((N - 1)!)^(-1)
             * (N-1!)^(-1) * N -> ((N - 2)!)^(-1)
             */
            reverseFactorial[i] = (reverseFactorial[i + 1] * (i + 1)) % mod;
        }
        //		reverseFactorial[0] = 1;
        //		for (int i = 0; i < FACTORIAL_NUM; i++) {
        //			reverseFactorial[i] = modInverse(factorial[i], mod);
        //		}
    }

    /**
     * フェルマーの小定理を用いて逆元を求める。
     * ある数xのmod p（pは素数）の逆数x'はx' = x^(p-2)
     * 繰り返し二乗法を用いて計算する。
     * http://satanic0258.hatenablog.com/entry/2016/04/29/004730
     * {@link BigInteger#modInverse(BigInteger)}とどちらが速いか？
     *
     * @param x
     * @return
     */
    private long modInverse(long x, long mod) {
        long res = 1L;
        long k = mod - 2L;
        long y = x;
        while (k != 0) {
            if (k % 2 != 0) {
                res = (res * y) % mod;
            }
            y = (y * y) % mod;
            k /= 2;
        }
        return res;
    }

    private void initFactorialCalc(long mod) {
        if (isInitFactorial) {
            //初期化済みならreturn
            return;
        }
        //階乗の事前計算
        initFactorial(mod);
        //逆数の計算もしておく
        initReverseFactorial(mod);
        isInitFactorial = true;
    }

    /**
     * 組み合わせ
     * nCk -> n! / k!・(n-k)!
     *
     * @param n
     * @param k
     * @param mod
     * @return
     */
    private long combination(int n, int k, long mod) {
        initFactorialCalc(mod);
        if (k < 0 || n < k) {
            return 0;
        }

        //分子の計算
        long calResN = factorial[n];

        //分母の計算
        long calResNK = reverseFactorial[n - k];
        long calcResK = reverseFactorial[k];
        long denominator = (calResNK * calcResK) % mod;

        return (calResN * denominator) % mod;

    }

    /**
     * 順列
     * nPk -> n! / (n-k)!
     *
     * @param n
     * @param k
     * @return
     */
    private long permutation(int n, int k, long mod) {
        initFactorialCalc(mod);
        if (k < 0 || n < k) {
            return 0;
        }
        long calResN = factorial[n];
        long calResNK = reverseFactorial[n - k];
        return (calResN * calResNK) % mod;
    }

    /**
     * 多項係数
     * 文字aをp個,bをq個,cをr個, dをs個 あわせてn個を1列に並べるときの並べ方
     * n! / p!・q!・r!・s!
     *
     * @param n
     * @param strNum
     * @param mod
     * @return
     */
    private long multinominalCoefficient(int n, List<Integer> strNum, int mod) {
        initFactorialCalc(mod);

        long calResN = factorial[n];
        long calResDenomi = 1;
        for (Integer num : strNum) {
            calResDenomi *= reverseFactorial[num];
        }
        return (calResN * calResDenomi) % mod;
    }

    /**
     * 重複組合せ nHr （同次積）
     * nHr = (n+r-1)Cr
     * 異なるn個のものから重複を許してr個取る組合せの総数
     * 例：
     * リンゴ，ミカン，牛肉の3種類の果物があります．これらの中から6個の食材を買って帰ります．
     * このとき，何通りの買い方がありますか？ただし，含まれない食材があってもよいものとします
     *
     * @param n
     * @param r
     * @param mod
     * @return
     */
    private long homogeneousProduct(int n, int r, long mod) {
        if (n == 0 && r == 0) {
            return 1;
        }
        return combination(n + r - 1, r, mod);
    }

    private final PrintWriter out = new PrintWriter(System.out);
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }

    private int readByte() {
        if (hasNextByte())
            return buffer[ptr++];
        else
            return -1;
    }

    private static boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }

    private void skipUnprintable() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr]))
            ptr++;
    }

    public boolean hasNext() {
        skipUnprintable();
        return hasNextByte();
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public String next() {
        if (!hasNext())
            throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }

    public long nextLong() {
        if (!hasNext())
            throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if (b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
}