백준 알고리즘 16173번 : 점프왕 쩰리 (Small)

링크

https://www.acmicpc.net/problem/16173

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;

wector<int> adj(3,vector<int>(3,0));
wector<int> dist(3,vector<int>(3,-1));
int dx[] = {1,0};
int dy[] = {0,1};

int main() {
  fastio;
  int n; cin >> n;
  for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) cin >> adj[i][j];
  queue<pii> Q;
  dist[0][0] = 0;
  Q.push({0, 0});
  while(!Q.empty()){
    auto [curX, curY] = Q.front(); Q.pop();
    int dash = adj[curX][curY];
    for(int i = 0; i < 2; i++){
      auto nx = curX + (dx[i] * dash);
      auto ny = curY + (dy[i] * dash);
      if(nx < 0 || nx >= n || ny < 0 || ny >= n) continue;
      if(dist[nx][ny] != -1) continue;
      dist[nx][ny] = dist[curX][curY] + 1;
      Q.push({nx, ny});
    }
  }
  cout << (dist[n - 1][n - 1]==-1 ? "Hing" : "HaruHaru") << "\n";
  return 0;
}

범위가 작아서 브루트하게 돌려줘도 되지만 BFS를 활용했습니다.
탐색을 할 때 오른쪽과 아래밖에 안되며, 각 칸의 수만큼 이동할 수 있다고 하였으니 그 부분에 대한 처리가 필요합니다.