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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define SZ(x) ((int)(x).size())
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define REP1(i,a,b) for (int i = a; i <= (int)(b); i++)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define PB push_back
#define MP make_pair
using namespace std;
typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;

template<class T> inline bool chmax( T &a, const T &b ) { return b>a ? a=b,true : false; }
template<class T> inline bool chmin( T &a, const T &b ) { return b<a ? a=b,true : false; }

struct UnionFind
{
  int n, num;
  vector<int> sz, par;
  UnionFind() {}
  UnionFind(int n_) : n(n_), num(n_), sz(n_, 1), par(n_, 0) { iota(par.begin(), par.end(), 0); }
  int find(int x)
  {
    return (x == par[x] ? x : par[x] = find(par[x]));
  }
  bool same(int x, int y)
  {
    return find(x) == find(y);
  }
  void unite(int x, int y)
  {
    x = find(x);
    y = find(y);
    if (x == y)
      return;
    if (sz[x] < sz[y])
      swap(x, y);
    sz[x] += sz[y];
    par[y] = x;
    num--;
  }
  int size(int x)
  {
    return sz[find(x)];
  }
  int count() const
  {
    return num;
  }
};
//

int h, w;
vector<string> in;

int f(int a, int b) {
  return a*w+b;
}

bool check( int a, int b, int c, int d ) {
    return (in[a][b] == '#' && in[c][d] == '.') || (in[a][b] == '.' && in[c][d] == '#');
}

int main() {
  cin >> h >> w;
  in.resize(h);
  UnionFind uf(h*w);
  REP(i,h) cin >> in[i];
  REP(i,h) REP(j,w) {
    if ( i>0 ) {
      if ( check(i,j,i-1,j) )
        uf.unite(f(i, j), f(i - 1, j));
    }
    if ( j>0 ) {
      if (check(i, j, i, j-1))
        uf.unite(f(i,j), f(i,j-1));
    }
  }
  VI B(h*w),W(h*w);
  REP(i,h) REP(j,w) {
    int p=uf.find(f(i,j));
    if ( in[i][j]=='.' ) W[p]++;
    else B[p]++;
  }
  int ans=0;
  REP(i,h) REP(j,w) if ( f(i,j)==uf.find(f(i,j)) ) {
    ans+=B[f(i,j)]*W[f(i,j)];
  }
  cout << ans << '\n';
  return 0;
}