力扣695. 岛屿的最大面积

 BFS写法：

class Solution {
public:
    typedef pair<int, int> pii;
    map<pii, bool> visited;
    vector<vector<int>> grid;
    int orientation[8] = {1, 0, -1, 0, 0, 1, 0 ,-1};
    int n, m;
    int result = 0;
    void bfs(vector<pii> nxt, int area) {
        if (nxt.empty()) {
            result = max(result, area);
            return;
        }
        vector<pii> v; //此处最好使用全局的双端队列，可以节省很多空间
        for (auto i : nxt) {
            for (int j = 0; j < size(orientation) / 2; ++j) {
                int x = i.first + orientation[j * 2], y = i.second + orientation[j * 2 + 1];
                if (x >= 0 && x < n && y >= 0 && y < m) {
                    pii temp = make_pair(x, y);
                    if (grid[x][y] == 1 && !visited[temp]) {
                        v.push_back(temp);
                        visited[temp] = true;
                        area++;
                    }
                }
            }
        }
        bfs(v, area);
    }
    int maxAreaOfIsland(vector<vector<int>>& grid) {
         this -> grid = grid;
         n = grid.size();
         m = grid[0].size();
         for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                pii temp = make_pair(i, j);
                if (grid[i][j] == 1 && !visited[temp]) {
                    visited[temp] = true;
                    bfs({temp}, 1);
                }
            }
         }
         return result;
    }
};