백준 알고리즘 13565번 : 침투

링크

https://www.acmicpc.net/problem/13565

sol1) DFS

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define X first
#define Y second
#define pb push_back
#define fastio cin.tie(0)->sync_with_stdio(0)
#define MAX(a, b) (((a) > (b)) ? (a) : (b))
#define sz(v) (int)(v).size()
#define all(v) v.begin(), v.end()
#define rall(v) (v).rbegin(), (v).rend()
#define compress(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define OOB(x, y) ((x) < 0 || (x) >= n || (y) < 0 || (y) >= m)
#define debug(x) cout << (#x) << ": " << (x) << '\n'
using namespace std;
using ll = long long;
using ull = unsigned long long;
using dbl = double;
using ldb = long double;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vs = vector<string>;
using tii = tuple<int,int,int>;
template<typename T> using wector = vector<vector<T>>;

const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const ll LNF = 1e18 + 7;

int m,n;
bool vist[1001][1001];
int adj[1001][1001];

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

void dfs(int x,int y){
  vist[x][y] = true;
  for(int i = 0; i < 4; i++){
    auto nx = x + dx[i];
    auto ny = y + dy[i];
    if(vist[nx][ny] || adj[nx][ny]) continue;
    if(nx < 0 || nx >= m || ny < 0 || ny >= n) continue;
    vist[nx][ny] = true;
    dfs(nx, ny);
  }
}

int main() {
	//fastio;
  cin >> m >> n;
  for(int i = 0; i < m; i++){
    for(int j = 0; j < n; j++){
      scanf("%1d",&adj[i][j]);
    }
  }
  for(int j = 0; j < n; j++){
    if(!adj[0][j] && vist[0][j] == false) dfs(0,j);
  }
  bool flag = false;
  for(int i = 0; i < n; i++){
    if(vist[m - 1][i]) flag = true;
  }
  cout << (flag ? "YES" : "NO") << "\n";
	return 0;
}

DFS를 이용해서 끝까지 갈 수 있는지 확인해주면 됩니다.
DFS를 돌린 후 맨 마지막줄의 vist배열만 확인해서 방문했다면 YES를 아니라면 NO를 출력해주면 됩니다.