For this problem, we're given a grid and want to connect all the interiors to each other. If we think of each interior region as a point and add edges between points representing the length of the fence that separates them, then this problem becomes one of finding a minimum spanning tree on a graph. Since our graph has $O(n*m)$ vertices and each vertex has at most 4 edges, it also has $O(n*m)$ edges, meaning that we can simply run our favorite minimum spanning tree algorithm to solve the problem.

However, we can actually make this solution even faster! If you're interested, see the platinum version of this problem.

Here is Mark Gordon's code, with an implementation of Kruskal's algorithm:

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

int A, B, N, M;

int nd(int r, int c) {
  return r * M + c;
}

int P[2010*2010];

int find(int x) {
  return x == P[x] ? x : P[x] = find(P[x]);
}

bool merge(int x, int y) {
  int X = find(x);
  int Y = find(y);
  if (X == Y) return false;
  P[X] = P[Y] = P[x] = P[y] = rand() % 2 ? X : Y;
  return true;
}

int main() {
  cin >> A >> B >> N >> M;

  vector<int> VA(N + 1), HA(M + 1);
  for (int i = 0; i < N; i++) {
    cin >> VA[i];
  }
  for (int j = 0; j < M; j++) {
    cin >> HA[j];
  }

  sort(VA.begin(), VA.end());
  for (int i = 0; i < N; i++) {
    VA[i] = VA[i + 1] - VA[i];
  }
  VA[N] = B - VA[N];

  sort(HA.begin(), HA.end());
  for (int i = 0; i < M; i++) {
    HA[i] = HA[i + 1] - HA[i];
  }
  HA[M] = A - HA[M];

  sort(VA.begin(), VA.end());
  sort(HA.begin(), HA.end());
  N++; M++;

  for (int i = 0; i < N * M; i++) {
    P[i] = i;
  }

  long long result = 0;
  for (int vi = 0, hi = 0; vi < N || hi < M; ) {
    if (hi == M || (vi < N && VA[vi] < HA[hi])) {
      for (int i = 0; i + 1 < M; i++) {
        if (merge(nd(vi, i), nd(vi, i + 1))) {
          result += VA[vi];
        }
      }
      vi++;
    } else {
      for (int i = 0; i + 1 < N; i++) {
        if (merge(nd(i, hi), nd(i + 1, hi))) {
          result += HA[hi];
        }
      }
      hi++;
    }
  }
  cout << result << endl;

  return 0;
}