LeetCode 684. 冗余连接 | 1627. 带阈值的图连通性（并查集）

LeetCode 684. 冗余连接

 简单并查集, 搞出一棵树即可

class Solution {
public:
    int fi(int x, vector<int>& fa) {
        return x == fa[x] ? x : fa[x] = fi(fa[x], fa);
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        vector<int> ans;
        vector<int> fa(n + 1, 0);
        for (int i = 0; i <= n; i++) fa[i] = i;
        for (auto p : edges) {
            int x = fi(p[0], fa), y = fi(p[1], fa);
            if (x != y) {
                fa[x] = y;
            } else {
                ans = p;
                break;
            }
        }
        return ans;
    }
};

LeetCode 1627. 带阈值的图连通性

A和B, P为A,B的公因子, 连接P和A即可, 最后再判断,  用并查集处理即可

class Solution {
public:
    int fi(int x, vector<int>& fa) {
        return x == fa[x] ? x : fa[x] = fi(fa[x], fa);
    }
    vector<bool> areConnected(int n, int threshold, vector<vector<int>>& queries) {
        vector<bool> ans;
        vector<int> fa(n + 1, 0);
        for (int i = 1; i <= n; i++) fa[i] = i;
        for (int i = threshold + 1; i <= n; i++) {
            for (int j = 2 * i; j <= n; j += i) {
                int fx = fi(i, fa), fy = fi(j, fa);
                if (fx != fy) fa[fx] = fy;    
            }
        }
        int m = queries.size();
        for (int i = 0; i < m; i++) {
            int fx = fi(queries[i][0], fa), fy = fi(queries[i][1], fa);
            ans.push_back(fx == fy);
        }
        return ans;
    }
};