Number of Enclaves

https://leetcode.com/contest/weekly-contest-130/problems/number-of-enclaves/

Given a 2D arrayA, each cell is 0 (representing sea) or 1 (representing land)

A move consists of walking from one land square 4-directionally to another land square, or off the boundary of the grid.

Return the number of land squares in the grid for which wecannotwalk off the boundary of the grid in any number of moves.

Example 1:

Input: 
[[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
Output: 
3
Explanation: 

There are three 1s that are enclosed by 0s, and one 1 that isn't enclosed because its on the boundary.

Example 2:

Input: 
[[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
Output: 
0
Explanation: 

All 1s are either on the boundary or can reach the boundary

Solution

每个相连组成的1的块是一个集合，也就是看集合中有没有在边界的。有关“构成集合”的题无非UF和DFS。

class Solution {
public:
    int numEnclaves(vector<vector<int>>& A) {
        vector<pair<int, int>> dirs = {{0, 1}, {1, 0}};
        int m = A.size(), n = A[0].size();
        for (int i = 0; i < m * n; ++i) parents.push_back(i);
        set<int> bos;
        int total = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (A[i][j] == 1) {
                    int ind = i * n + j;
                    ++total;
                    if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                        bos.insert(ind);
                    }
                    for (const auto dir : dirs) {
                        int x = i + dir.first;
                        int y = j + dir.second;
                        if (x >= 0 && x < m && y >= 0 && y < n && A[x][y] == 1) {
                            un(x * n + y, ind);
                        }
                    }
                }
            }
        }

        for (int b : bos) {
            bos.insert(find(b));
        }

        for (int p : parents) {
            if (bos.find(find(p)) != bos.end()) --total;
        }
        return total;
    }

private:
    vector<int> parents;

    int td21d(int x, int y, int n) {
        return x * n + y;
    }

    int find(int x) {
        if (x != parents[x]) {
            parents[x] = find(parents[x]);
        }
        return parents[x];
    }

    void un(int x, int y) {
        x = find(x), y = find(y);
        if (x != y) {
            parents[x] = y;
        }
    }
};