leetcode1905 统计子岛屿

思路：

在grid2中bfs/dfs寻找岛屿的同时判断这个岛屿对应的位置在grid1中是否全部为1。

实现：

class Solution
{
public:
    int dx[4] = {0, 1, 0, -1};
    int dy[4] = {1, 0, -1, 0};
    bool dfs(int x, int y, vector<vector<int>>& g1, vector<vector<int>>& g2, vector<vector<bool>>& vis)
    {
        bool res = true;
        if (g1[x][y] == 0) res = false; 
        vis[x][y] = true;
        int n = g2.size(), m = g2[0].size();
        for (int i = 0; i < 4; i++)
        {
            int nx = x + dx[i], ny = y + dy[i];
            if (nx >= 0 and nx < n and ny >= 0 and ny < m and g2[nx][ny] == 1 and !vis[nx][ny])
            {
                res &= dfs(nx, ny, g1, g2, vis);
            }
        }
        return res;
    }
    int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2)
    {
        int n = grid1.size(), m = grid1[0].size();
        vector<vector<bool>> vis(n, vector<bool>(m, false));
        int res = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (grid2[i][j] == 1 and !vis[i][j])
                {
                    res += dfs(i, j, grid1, grid2, vis);
                }
            }
        }
        return res;
    }
}