Problem:
Score : 300 points

Problem Statement

There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W.
If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is #; if that square is painted white, the j-th character in the string S_i is ..

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

1 \leq H, W \leq 400

|S_i| = W (1 \leq i \leq H)

For each i (1 \leq i \leq H), the string S_i consists of characters # and ..

Input

Input is given from Standard Input in the following format:

H W
S_1
S_2
:
S_H

Output

Print the answer.

Sample Input 1

3 3
.#.
..#
#..

Sample Output 1

10

Some of the pairs satisfying the condition are ((1, 2), (3, 3)) and ((3, 1), (3, 2)), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

Sample Input 2

2 4
....
....

Sample Output 2

0

Sample Input 3

4 3
###
###
...
###

Sample Output 3

6

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using ll  = int64_t;
using P   = pair<ll, ll>;
using mat = vector<vector<ll>>;

void IOS() { ios::sync_with_stdio(false), cin.tie(0); }

template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    fill((T *)array, (T *)(array + N), val);
}

const ll INF = 1e16;
const ll MOD = 1e9 + 7;
int dx[4]    = {1, 0, -1, 0};
int dy[4]    = {0, 1, 0, -1};
// ll pas[5001][5002];
ll h, w;
string s[402];
ll dp[402][402][2]  = {};
bool used[402][402] = {};
ll rec(int i, int j, bool isW) {
    if (dp[i][j][isW] > 0) return dp[i][j][isW];
    // cout << i << " " << j << endl;
    used[i][j] = true;
    char p     = s[i][j];
    ll &tmp    = dp[i][j][isW];
    tmp        = (isW && s[i][j] == '#') || (!isW && s[i][j] != '#');
    int ni = i + 1, nj = j;
    if (ni < h && s[ni][nj] != p && !used[ni][nj]) tmp += rec(ni, nj, isW);
    ni = i, nj = j + 1;
    if (nj < w && s[ni][nj] != p && !used[ni][nj]) tmp += rec(ni, nj, isW);

    return tmp;
}

signed main() {
    IOS();
    cin >> h >> w;
    for (int i = 0; i < h; ++i) { cin >> s[i]; }
    ll ans = 0;
    for (int i = 0; i < h; ++i) {
        for (int j = 0; j < w; ++j) {
            ans += rec(i, j, s[i][j] == '.');
            Fill(used, false);
        }
    }
    cout << ans << endl;

    return 0;
}