分析

• 对于某一个方向是否有墙的判断：可以用二进制判断，左上右下分别用0、1、2、3代表，则可以使用g[x][y] >> i & 1判断（这里的i就是刚才的0、1、2、3）。其他直接使用floodfill算法求解即可。

BFS

// BFS
#include <iostream>

#define x first
#define y second

using namespace std;

typedef pair<int, int> PII;

const int N = 55, M = N * N;

int n, m;
int g[N][N];
PII q[M];
bool st[N][N];

int bfs(int sx, int sy) {

    int hh = 0, tt = 0;
    int area = 0;

    int dx[4] = {0, -1, 0, 1}, dy[4] = {-1, 0, 1, 0};  // 左上右下

    q[0] = {sx, sy};
    st[sx][sy] = true;

    while (hh <= tt) {
        auto t = q[hh++];
        area++;

        for (int i = 0; i < 4; i++) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if (st[a][b]) continue;
            if (g[t.x][t.y] >> i & 1) continue;

            q[++tt] = {a, b};
            st[a][b] = true;
        }
    }
}

int main() {

    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> g[i][j];

    int cnt = 0, area = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (!st[i][j]) {
                area = max(area, bfs(i, j));
                cnt++;
            }

    cout << cnt << endl;
    cout << area << endl;

    return 0;
}

DFS

// DFS
#include <iostream>

#define x first
#define y second

using namespace std;

const int N = 55;

int n, m;
int g[N][N];
bool st[N][N];
int dx[] = {0, -1, 0, 1}, dy[] = {-1, 0, 1, 0};  // 上右下左

int dfs(int sx, int sy) {

    st[sx][sy] = true;
    int area = 1;
    for (int i = 0; i < 4; i++) {
        int a = sx + dx[i], b = sy + dy[i];
        if (a >= 0 && a < n && b >= 0 && b < m && !(g[sx][sy] >> i & 1) && !st[a][b])
            area += dfs(a, b);
    }
    return area;
}

int main() {

    cin >> n >> m;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cin >> g[i][j];

    int cnt = 0, area = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            if (!st[i][j]) {
                area = max(area, dfs(i, j));
                cnt++;
            }

    cout << cnt << endl;
    cout << area << endl;

    return 0;
}