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15092번: Sheba’s Amoebas ↗

Solutions

C++14
2.5 KB | 2543 chars
/*
[15092: Sheba’s Amoebas](https://www.acmicpc.net/problem/15092)

Tier: Silver 2 
Category: graphs, graph_traversal, bfs, dfs
*/

#include <bits/stdc++.h>

using namespace std;

#define for1(s, e) for(int i = s; i < e; i++)
#define for1j(s, e) for(int j = s; j < e; j++)
#define forEach(k) for(auto i : k)
#define forEachj(k) for(auto j : k)
#define sz(vct) vct.size()
#define all(vct) vct.begin(), vct.end()
#define sortv(vct) sort(vct.begin(), vct.end())
#define uniq(vct) sort(all(vct));vct.erase(unique(all(vct)), vct.end())
#define fi first
#define se second
#define INF (1ll << 60ll)

typedef unsigned long long ull;
typedef long long ll;
typedef ll llint;
typedef unsigned int uint;
typedef unsigned long long int ull;
typedef ull ullint;

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef pair<double, int> pdi;
typedef pair<string, string> pss;

typedef vector<int> iv1;
typedef vector<iv1> iv2;
typedef vector<ll> llv1;
typedef vector<llv1> llv2;

typedef vector<pii> piiv1;
typedef vector<piiv1> piiv2;
typedef vector<pll> pllv1;
typedef vector<pllv1> pllv2;
typedef vector<pdd> pddv1;
typedef vector<pddv1> pddv2;

const double EPS = 1e-8;
const double PI = acos(-1);

template<typename T>
T sq(T x) { return x * x; }

int sign(ll x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(int x) { return x < 0 ? -1 : x > 0 ? 1 : 0; }
int sign(double x) { return abs(x) < EPS ? 0 : x < 0 ? -1 : 1; }

int n, m;
vector<string> grid;
vector<vector<bool>> visited;

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

void solve() {
  cin >> n >> m;
  grid.resize(n);
  for1(0, n) {
    cin >> grid[i];
  }

  visited.resize(n, vector<bool>(m, false));

  int ans = 0;

  for1(0, n) {
    for1j(0, m) {
      if(grid[i][j] != '#') continue;
      if(visited[i][j]) continue;

      ans++;
      queue<pii> q;
      q.push({ i, j });

      while(!q.empty()) {
        pii cur = q.front();
        q.pop();

        if(visited[cur.fi][cur.se]) continue;
        visited[cur.fi][cur.se] = true;

        for(int d = 0; d < 8; d++) {
          int ny = cur.fi + dy[d];
          int nx = cur.se + dx[d];

          if(ny < 0 || ny >= n || nx < 0 || nx >= m) continue;
          if(grid[ny][nx] != '#') continue;
          if(visited[ny][nx]) continue;

          q.push({ ny, nx });
        }
      }
    }
  }

  cout << ans << '\n';
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);cout.tie(NULL);
  int tc = 1; // cin >> tc;
  while(tc--) solve();

}