原题链接 https://leetcode.com/problems/largest-component-size-by-common-factor/submissions/

题意就不再做解释了，直接上题解：

方法一：简单并查集--超时

        只要任何两个数之间有相同的因子，那么就把他们连接到一起，这样可以把每个数字和它的所有因子进行链接，最后统计哪个因子上面的数字最多即可，代码比较好懂

class Solution {
    vector<int>father;
    int findfather(int n){
        while(n!=father[n]){
            n = father[n];
        }
        return n;
    }
    void merge(int a,int b){
        int aa = findfather(a);
        int bb = findfather(b);
        if(aa!=bb)
            father[aa]=bb;
    }
public:
    int largestComponentSize(vector<int>& A) {
        int maxnum = 0;
        for(auto x:A)
            maxnum = max(maxnum,x);
        father = vector<int>(maxnum+1,-1);
        for(int i = 0;i<maxnum+1;i++)
            father[i] = i;

        int n = A.size();
        for(auto x:A)
            for(int i = 2;i<=sqrt(x);i++)
                if(x%i==0){
                    merge(x,i);
                    merge(x,x/i);
                }

        unordered_map<int,int>mp;
        for(auto x:A){
            // cout<<father[x]<<endl;
            mp[findfather(x)]++;
        }
        int res = 0;
        for(auto x:mp)
            res = max(res,x.second);
        return res;
        
    }
};

方法二

        然鹅却超时了，怀疑是因为数组太长，寻根部分花了较多时间，于是考虑对每个数字分别分解质因数，开一个二维 vector 记录每个质因数所对应的数字有哪些，然后将其记录的所有数字通过并查集连接在一起，最后的结果是最大的集合，代码参考：https://www.acwing.com/solution/LeetCode/content/646/

class Solution {
public:
    vector<int> fa, sz;

    int find(int x) {
        return x != fa[x] ? fa[x] = find(fa[x]) : x;
    }

    int largestComponentSize(vector<int>& A) {
        int n = A.size(), ans = 0;
        vector<vector<int>> num(100001);

        for (int i = 0; i < n; i++) {
            int x = A[i];
            for (int y = 2; y * y <= x; y++) {
                if (x % y == 0)
                    num[y].push_back(i);
                while (x % y == 0)
                    x /= y;
            }
            if (x > 1)
                num[x].push_back(i);
        }

        fa = vector<int>(n);
        sz = vector<int>(n);

        for (int i = 0; i < n; i++) {
            fa[i] = i;
            sz[i] = 1;
        }

        for (int i = 1; i <= 100000; i++)
            for (int j = 1; j < num[i].size(); j++) {
                int x = num[i][0], y = num[i][j];
                int fx = find(x), fy = find(y);
                if (fx != fy) {
                    if (sz[fx] < sz[fy]) {
                        fa[fy] = fx;
                        sz[fx] += sz[fy];
                    }
                    else {
                        fa[fx] = fy;
                        sz[fy] += sz[fx];
                    }
                }
            }

        for (int i = 0; i < n; i++)
            if (i == find(i))
                ans = max(ans, sz[i]);

        return ans;
    }
};