岛屿的最大面积

. - 力扣（LeetCode）

1.刚开始mn又加了int

 2.bfs里符合条件了，不push，，，在写什么几把

class Solution {
public:
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    bool vis[50][50];
    int m, n;
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        m = grid.size(), n = grid[0].size();
        int ret = 0;
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid[i][j] && !vis[i][j])
                {
                    ret = max(ret, bfs(grid, i, j));
                }
            }
        }
        return ret;
    }
    int bfs(vector<vector<int>>& grid, int i, int j)
    {
        queue<pair<int, int>> q;
        q.push({i, j});
        vis[i][j] = true;
        int count = 1;
        while(q.size())
        {
            auto e = q.front(); q.pop();
            for(int i = 0; i < 4; i++)
            {
                int x = e.first + dx[i], y = e.second + dy[i];
                if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] && !vis[x][y])
                {
                    count++;
                    q.push({x, y});
                    vis[x][y] = true;
                }
            }
        }
        return count;
    }
};

 被围绕的区域

. - 力扣（LeetCode）

 就是需要对外圈进行标记，一种是改变原来的值，最后在变回来，另一种是vis进行标记，

class Solution {
public:
    
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
    void solve(vector<vector<char>>& board) {
        m = board.size(), n = board[0].size();
        // for(int i = 0; i < n; i++)
        //     dfs(board, 0, i), dfs(board, m - 1, i);
        // for(int i = 0; i < m; i++)
        //     dfs(board, i, 0), dfs(board, i, n - 1);
        for(int i = 0; i < n; i++)
        {
            if(board[0][i] == 'O') bfs(board, 0, i);
            if(board[m - 1][i] == 'O') bfs(board, m - 1, i);
        }
        for(int i = 0; i < m; i++)
        {
            if(board[i][0] == 'O') bfs(board, i, 0);
            if(board[i][n - 1] == 'O') bfs(board, i, n - 1);
        }
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(board[i][j] == 'O') board[i][j] = 'X';
                else if(board[i][j] == ' ') board[i][j] = 'O';
            }
        }
    }

    void bfs(vector<vector<char>>& board, int i, int j)
    {
        queue<pair<int, int>> q;
        q.push({i, j});
        board[i][j] = ' ';
        while(q.size())
        {
            auto e = q.front(); q.pop();
            for(int i = 0; i < 4; i++)
            {
                int x = e.first + dx[i], y = e.second + dy[i];
                if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
                {
                    board[x][y] = ' ';
                    q.push({x, y});
                }
            }
        }
    }
};