[leetcode] 827. 最大人工岛 | 二维并查集

Posted 2023-02-20 PushyTao

class Solution 
private:
    int size[507 * 507];
    int fa[507 * 507];
    void _init() 
        for(int i=0;i<=n * m;i++) fa[i] = i, size[i] = 1;
    
    int _find(int x) 
        if(fa[x] == x) return x;
        else return fa[x] = _find(fa[x]);
    
    int maxSize = 1;
    void _union(int u,int v) 
        int fau = _find(u);
        int fav = _find(v);
        if(fau == fav) return ;
        fa[fau] = fav;
        size[fav] += size[fau];
        maxSize = max(maxSize, size[fav]);
    
    int n, m;
    int get(int x, int y) 
        return x * n + y;
    
    bool edge(int x, int y) 
        if(x < 0 || x >= n || y < 0 || y >= m) return false;
        return true;
    
public:
    int largestIsland(vector<vector<int>>& grid) 
        n = grid.size();
        m = grid[0].size();

        _init();

        int dx[4] = 0, 0, -1, 1;
        int dy[4] = -1, 1, 0, 0;
        
        for(int i = 0; i < n; i ++) 
            for(int j = 0; j < m; j ++) 
                if(!grid[i][j]) continue;
                for(int to = 0; to < 4; to ++) 
                    int tx = i + dx[to];
                    int ty = j + dy[to];
                    if(edge(tx, ty) && grid[tx][ty]) 
                        _union(get(i,j), get(tx,ty));
                    
                
            
        
        
        unordered_map<int,int> mp;
        for(int i = 0;i < n; i ++) 
            for(int j = 0; j < m; j ++) 
                if(grid[i][j]) continue;
                for(int to = 0; to < 4; to ++) 
                    int tx = i + dx[to];
                    
                    int ty = j + dy[to];
                    if(!edge(tx, ty)) continue;
                    if(grid[tx][ty] == 0) continue;
                    
                    int fa = _find(get(tx, ty));
                    mp[fa] = size[fa];
                
                int total = 1;
                for(unordered_map<int,int> ::iterator it = mp.begin(); it != mp.end(); it ++) 
                    total += it->second;
                
                //for(auto at: mp) total += at.second;
                maxSize = max(maxSize, total);
                mp.clear();
            
        

        return maxSize;
    
;