Problem p02735 (C++)

Buggy submission (status: Time Limit Exceeded):
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;
using graph = vector<vector<int>>;
using pii = pair<int, int>;

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repd(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= (ll)(b); i--)
// xにはvectorなどのコンテナ
#define all(x) (x).begin(), (x).end()
#define size(x) ((ll)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

#define INF 1000000000000
#define M 10000007
#define MAXR 100000

#define pb push_back
#define mp make_pair
#define f first
#define s second
#define FIND(x, y) find(all(x), y) != x.end()
template <class T> inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T> inline bool chmin(T &a, T b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}

struct uf {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
  vector<int> rcount;

  uf(int N) : par(N) { // 最初は全てが根であるとして初期化
    for (int i = 0; i < N; i++)
      par[i] = i;
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x)
      return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x);        // xの根をrx
    int ry = root(y);        // yの根をry
    if (rx == ry)
      return; // xとyの根が同じ(=同じ木にある)時はそのまま
    par[rx] =
        ry; // xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
};
template <int MOD> struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
    if (val < 0)
      val += MOD;
  }
  constexpr int getmod() { return MOD; }
  constexpr Fp operator-() const noexcept { return val ? MOD - val : 0; }
  constexpr Fp operator+(const Fp &r) const noexcept { return Fp(*this) += r; }
  constexpr Fp operator-(const Fp &r) const noexcept { return Fp(*this) -= r; }
  constexpr Fp operator*(const Fp &r) const noexcept { return Fp(*this) *= r; }
  constexpr Fp operator/(const Fp &r) const noexcept { return Fp(*this) /= r; }
  constexpr Fp &operator+=(const Fp &r) noexcept {
    val += r.val;
    if (val >= MOD)
      val -= MOD;
    return *this;
  }
  constexpr Fp &operator-=(const Fp &r) noexcept {
    val -= r.val;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr Fp &operator*=(const Fp &r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp &operator/=(const Fp &r) noexcept {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp &r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp &r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
    if (n == 0)
      return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1)
      t = t * a;
    return t;
  }
};

const int MOD = 1000000007;
using mint = Fp<MOD>;

const int dx[2] = {1, 0};
const int dy[2] = {0, 1};

bool flag = false;

int H, W, ans = 10000000;
vector<string> field;

int seen[510][510];
void dfs(int h, int w, int gh, int gw) {
  rep(dir, 2) {
    flag = false;
    int nh = h + dy[dir];
    int nw = w + dx[dir];

    if (nh < 0 || nh >= H || nw < 0 || nw >= W)
      continue;
    if (field[h][w] != '#' && field[nh][nw] == '#')
      flag = chmin(seen[nh][nw], seen[h][w] + 1);
    else
      flag = chmin(seen[nh][nw], seen[h][w]);
    if (nh == gh && nw == gw)
      continue;

    dfs(nh, nw, gh, gw);
  }
}

int main() {
  cin >> H >> W;
  field.resize(H);
  rep(h, H) cin >> field[h];

  int sh = 0, sw = 0, gh = H - 1, gw = W - 1;
  rep(i, H) rep(j, W) seen[i][j] = 1000;
  seen[0][0] = 0;
  dfs(sh, sw, gh, gw);
  if (field[0][0] == '#')
    seen[gh][gw]++;
  cout << seen[gh][gw] << endl;
}


Fixed submission:
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;
using graph = vector<vector<int>>;
using pii = pair<int, int>;

#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repd(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= (ll)(b); i--)
// xにはvectorなどのコンテナ
#define all(x) (x).begin(), (x).end()
#define size(x) ((ll)(x).size())
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

#define INF 1000000000000
#define M 10000007
#define MAXR 100000

#define pb push_back
#define mp make_pair
#define f first
#define s second
#define FIND(x, y) find(all(x), y) != x.end()
template <class T> inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T> inline bool chmin(T &a, T b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}

struct uf {
  vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
  vector<int> rcount;

  uf(int N) : par(N) { // 最初は全てが根であるとして初期化
    for (int i = 0; i < N; i++)
      par[i] = i;
  }

  int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
    if (par[x] == x)
      return x;
    return par[x] = root(par[x]);
  }

  void unite(int x, int y) { // xとyの木を併合
    int rx = root(x);        // xの根をrx
    int ry = root(y);        // yの根をry
    if (rx == ry)
      return; // xとyの根が同じ(=同じ木にある)時はそのまま
    par[rx] =
        ry; // xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
  }

  bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
    int rx = root(x);
    int ry = root(y);
    return rx == ry;
  }
};
template <int MOD> struct Fp {
  long long val;
  constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
    if (val < 0)
      val += MOD;
  }
  constexpr int getmod() { return MOD; }
  constexpr Fp operator-() const noexcept { return val ? MOD - val : 0; }
  constexpr Fp operator+(const Fp &r) const noexcept { return Fp(*this) += r; }
  constexpr Fp operator-(const Fp &r) const noexcept { return Fp(*this) -= r; }
  constexpr Fp operator*(const Fp &r) const noexcept { return Fp(*this) *= r; }
  constexpr Fp operator/(const Fp &r) const noexcept { return Fp(*this) /= r; }
  constexpr Fp &operator+=(const Fp &r) noexcept {
    val += r.val;
    if (val >= MOD)
      val -= MOD;
    return *this;
  }
  constexpr Fp &operator-=(const Fp &r) noexcept {
    val -= r.val;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr Fp &operator*=(const Fp &r) noexcept {
    val = val * r.val % MOD;
    return *this;
  }
  constexpr Fp &operator/=(const Fp &r) noexcept {
    long long a = r.val, b = MOD, u = 1, v = 0;
    while (b) {
      long long t = a / b;
      a -= t * b;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    val = val * u % MOD;
    if (val < 0)
      val += MOD;
    return *this;
  }
  constexpr bool operator==(const Fp &r) const noexcept {
    return this->val == r.val;
  }
  constexpr bool operator!=(const Fp &r) const noexcept {
    return this->val != r.val;
  }
  friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept {
    return os << x.val;
  }
  friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
    if (n == 0)
      return 1;
    auto t = modpow(a, n / 2);
    t = t * t;
    if (n & 1)
      t = t * a;
    return t;
  }
};

const int MOD = 1000000007;
using mint = Fp<MOD>;

const int dx[2] = {1, 0};
const int dy[2] = {0, 1};

bool flag = false;

int H, W, ans = 10000000;
vector<string> field;

int seen[510][510];
void dfs(int h, int w, int gh, int gw) {
  rep(dir, 2) {
    flag = false;
    int nh = h + dy[dir];
    int nw = w + dx[dir];

    if (nh < 0 || nh >= H || nw < 0 || nw >= W)
      continue;
    if (field[h][w] != '#' && field[nh][nw] == '#')
      flag = chmin(seen[nh][nw], seen[h][w] + 1);
    else
      flag = chmin(seen[nh][nw], seen[h][w]);
    if (nh == gh && nw == gw)
      continue;
    if (!flag)
      continue;

    dfs(nh, nw, gh, gw);
  }
}

int main() {
  cin >> H >> W;
  field.resize(H);
  rep(h, H) cin >> field[h];

  int sh = 0, sw = 0, gh = H - 1, gw = W - 1;
  rep(i, H) rep(j, W) seen[i][j] = 1000;
  seen[0][0] = 0;
  dfs(sh, sw, gh, gw);
  if (field[0][0] == '#')
    seen[gh][gw]++;
  cout << seen[gh][gw] << endl;
}