<p>You are given an integer array of unique positive integers <code>nums</code>. Consider the following graph:</p>

<ul>
	<li>There are <code>nums.length</code> nodes, labeled <code>nums[0]</code> to <code>nums[nums.length - 1]</code>,</li>
	<li>There is an undirected edge between <code>nums[i]</code> and <code>nums[j]</code> if <code>nums[i]</code> and <code>nums[j]</code> share a common factor greater than <code>1</code>.</li>
</ul>

<p>Return <em>the size of the largest connected component in the graph</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex1.png" style="width: 500px; height: 97px;" />
<pre>
<strong>Input:</strong> nums = [4,6,15,35]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex2.png" style="width: 500px; height: 85px;" />
<pre>
<strong>Input:</strong> nums = [20,50,9,63]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/01/ex3.png" style="width: 500px; height: 260px;" />
<pre>
<strong>Input:</strong> nums = [2,3,6,7,4,12,21,39]
<strong>Output:</strong> 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    /*
    如果nums[i]與nums[j]之間有公因數>1的話，這兩個值中間有連接
    尋找連線數量的最大值
    ★先找質數 並把每一個質因數歸類到各質數內
    對每一個nums元素進行因式分解 時間複雜度 nlogn
    */
    int L = 100000;
    int Father[100005];
    int largestComponentSize(vector<int>& nums) {
        vector<int> primes = Eratosthenes(L);

        for(int i = 0; i < 100005; ++i)
            Father[i] = i;

        for(int i = 0; i < nums.size(); ++i)
        {
            int x = nums[i];
            for(int p : primes)
            {
                if(x < p)
                    break;
                if(x % p == 0)
                {
                    if(FindFather(nums[i]) != FindFather(p))
                        Union(nums[i], p);
                    while(x % p == 0)
                        x /= p;
                }
            }
        }

        unordered_map<int, int> map;
        for(int i = 0; i < nums.size(); ++i)
            map[FindFather(nums[i])]++;

        int ans = 0;
        for(auto m : map)
            ans = max(ans, m.second);
        return ans;
    }

    int FindFather(int x)
    {
        if(x != Father[x])
            Father[x] = FindFather(Father[x]);
        return Father[x];
    }    

    void Union(int x, int y)
    {
        x = Father[x];
        y = Father[y];
        if(x > y)
            Father[x] = y;
        else
            Father[y] = x;
    }

    // 獲得0~n內所有質數的模板
    vector<int> Eratosthenes(int n)
    {
        auto q=vector<int>(n+1,0);
        for (int i=2; i<=sqrt(n); i++)
        {
            if (q[i]==0)
            {
                int j=i*2;
                while (j<n)
                {
                    q[j]=1;
                    j+=i;
                }
            }
        }
        vector<int>primes;
        for (int i=2; i<=n; i++)
        {
            if (q[i]==0)
                primes.push_back(i);                
        }
        return primes;
    }
};
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