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6031번: Feeding Time ↗

Solutions

C++14
2.5 KB | 2528 chars
/*
[6031: Feeding Time](https://www.acmicpc.net/problem/6031)

Tier: Silver 1
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; }

ll R, C;
vector<string> grid;
vector<vector<bool>> chk;
ll ans;

ll dy[] = { 0, 0, 1, -1, 1, 1, -1, -1};
ll dx[] = { 1, -1, 0, 0, 1, -1, 1, -1};
void solve() {
  cin >> C >> R;

  grid.resize(R);

  for1(0, R) cin >> grid[i];

  chk = vector<vector<bool>>(R, vector<bool>(C, 0));

  for1(0, R) {
    for1j(0, C) {
      if(grid[i][j] == '*') continue;
      if(chk[i][j]) continue;

      ll cnt = 0;
      queue <pll> Q;

      Q.push({i, j});

      while(!Q.empty()) {
        auto [y, x] = Q.front(); Q.pop();

        if(chk[y][x]) continue;
        chk[y][x] = true;
        cnt += 1;

        for(int d = 0; d < 8; d++) {
          ll ny = y + dy[d];
          ll nx = x + dx[d];

          if(ny < 0 || nx < 0 || ny >= R || nx >= C) continue;
          if(grid[ny][nx] == '*') continue;
          if(chk[ny][nx]) continue;
          
          Q.push({ny, nx});
        }
      }
      
      ans = max(ans, cnt);
    }
  }

  cout << ans;
}

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