2234번 성곽

오홍.. 안보고 잘했당ㅎㅎ
전 문제와 비슷한 면이 있어 잘 풀 수 있었다. 벽 부수고 이동하기랑도 비슷하다.

#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
using namespace std;

int dx[4] = { 0,-1,0,1 };
int dy[4] = { -1,0,1,0 };
int map[50][50];
int main() {
	freopen("in1.txt", "rt", stdin);
	int n, m;
	cin >> m >> n;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cin >> map[i][j];
		}
	}
	int num = -1;
	vector<int> p;
	vector<vector<int>> dist(n, vector<int>(m, -1));
	int ans = -1;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (dist[i][j] != -1) continue;
			num++;
			dist[i][j] = num;
			queue<pair<int, int>> Q;
			Q.push(make_pair(i, j));
			int cnt = 1;
			while (!Q.empty()) {
				int x, y;
				tie(x, y) = Q.front(); Q.pop();
				for (int i = 0; i < 4; i++) {
					int xx = x + dx[i];
					int yy = y + dy[i];
					if (xx >= n || xx < 0 || yy >= m || yy < 0) continue;
					if ((map[x][y] & 1 << i) == 0) {
						if (dist[xx][yy] == -1) {
							dist[xx][yy] = num;
							cnt++;
							Q.push(make_pair(xx, yy));
						}
					}
				}

			}
			p.push_back(cnt);
			if (cnt > ans) ans = cnt;
		}
	}
	cout << p.size() << '\n';
	cout << ans << '\n';
	//이제 dist배열을 이용하자
	ans = -1;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++ ) {
			for (int k = 0; k < 4; k++) {
				int xx = i + dx[k];
				int yy = j + dy[k];
				if (xx >= n || xx < 0 || yy >= m || yy < 0) continue;
				if (dist[i][j] == dist[xx][yy]) continue;
				if (map[i][j] & (1 << k)) {
					if (ans < p[dist[i][j]] + p[dist[xx][yy]]) {
						ans = p[dist[i][j]] + p[dist[xx][yy]];
					}
				}
			}
		}
	}
	cout << ans << '\n';
	return 0;

}