주난의 난 질문

저는 bfs와 dfs를 혼용하는 방식으로 코드를 짯는데 이렇게 짜면 단점이 어떤게 있을지 궁금합니다.

큰돌님 항상 강의 잘 듣고 있습니다 감사합니다.

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>

using namespace std;


int dx[4] = {0, 0, 1, -1};
int dy[4] = { 1, -1, 0, 0 };
int N, M, startX, startY, endX, endY;
char board[300][300] = { 0 };
int visit[300][300] = { 0 };
queue<pair<int, int>> q;


void print() {
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			cout << board[i][j] << " ";
		}
		cout << "\n";
	}
	cout << "\n";
}
void print2() {
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			cout << visit[i][j] << " ";
		}
		cout << "\n";
	}
	cout << "\n";
}

void dfs(int x, int y,  int a) {
	//print2();
	visit[y][x] = a;
	for (int i = 0; i < 4; i++) {
		int nx = x + dx[i];
		int ny = y + dy[i];
		if (nx <0 || nx > M || ny <0 || ny  > N)
			continue;
		if (!visit[ny][nx] && board[ny][nx] == '0') {
			dfs(nx, ny, a);
		}
		else if (!visit[ny][nx] && (board[ny][nx] == '1'|| board[ny][nx] == '#')) {
			board[ny][nx] = '0';
			visit[ny][nx] = a;
			q.push({ nx, ny });
		}
	}
}

void bfs() {
	q.push({ startX-1, startY-1 });
	visit[startY-1][startX-1] = 1;

	while (!q.empty()) {
		//print2();
		int x = q.front().first;
		int y = q.front().second;
		q.pop();
		for (int i = 0; i < 4; i++) {
			int nx = x + dx[i];
			int ny = y + dy[i];
			if (nx <0 || nx > M  || ny <0 || ny  > N)
				continue;
			if (!visit[ny][nx] && board[ny][nx] == '1') {
				q.push({ nx, ny });
				visit[ny][nx] = visit[y][x] + 1;
				board[ny][nx] = '0';
			}
			else if (!visit[ny][nx] && board[ny][nx] == '0') {
				dfs(nx, ny, visit[y][x] + 1);
				if (visit[endY-1][endX-1])
					return;
			}
			else if (!visit[ny][nx] && board[ny][nx] == '#') {
				visit[ny][nx] = visit[y][x] + 1; 
				return;
			}
		}
	}

}


int main() {
	char symbol;
	cin >> N >> M;
	cin >> startY >> startX >> endY >> endX;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			cin >> symbol;
			board[i][j] = symbol;
		}
	}
	bfs();
	
	cout << visit[endY - 1][endX - 1]-1;
}