백준 알고리즘 6146번 : 신아를 만나러

링크

https://www.acmicpc.net/problem/6146

sol1) BFS

#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 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 adj[1001][1001];
wector<int> dist(1001,vector<int>(1001, -1));
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};

int main() {
  fastio;
  int x,y,n; cin >> x >> y >> n;
  queue<pii> Q;
  for(int i = 0; i < n; i++){
    int a,b; cin >> a >> b;
    adj[a + 500][b + 500] = INF;
  }
  Q.push({500,500});
  dist[500][500] = 0;
  while(!Q.empty()){
    auto [curX,curY] = Q.front(); Q.pop();
    for(int i = 0; i < 4; i++){
      auto nx = curX + dx[i];
      auto ny = curY + dy[i];
      if(nx < 0 || nx > 1000 || ny < 0 || ny > 1000) continue;
      if(dist[nx][ny] != -1 || adj[nx][ny] == INF) continue;
      dist[nx][ny] = dist[curX][curY] + 1;
      Q.push({nx,ny});
    }
  }
  cout << dist[x + 500][y + 500] << "\n";
  return 0;
}

2차원 BFS문제였습니다.
물 웅덩이의 좌표가 -500~500으로 들어오기 때문에 1000짜리 배열을 만들고 입력받은 좌표마다 500을 더해준 후 BFS를 돌려주면 됩니다.