<p>On a 2D plane, we place <code>n</code> stones at some integer coordinate points. Each coordinate point may have at most one stone.</p>

<p>A stone can be removed if it shares either <strong>the same row or the same column</strong> as another stone that has not been removed.</p>

<p>Given an array <code>stones</code> of length <code>n</code> where <code>stones[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the location of the <code>i<sup>th</sup></code> stone, return <em>the largest possible number of stones that can be removed</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> [0,0] is the only stone on the plane, so you cannot remove it.
</pre>

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

<ul>
	<li><code>1 &lt;= stones.length &lt;= 1000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>No two stones are at the same coordinate point.</li>
</ul>

class Solution {
public:
    unordered_map<int,vector<int>> x;
    unordered_map<int,vector<int>> y;
    unordered_map<int,int> father;
    int n=10000;
    int find_father(int x){
        if(father[x]!=x){
            father[x]=find_father(father[x]);
        }
        return father[x];
    }
    void Union(int x, int y){
        x=father[x];
        y=father[y];
        if(x<=y){
            father[y]=x;
        }
        else{
            father[x]=y;
        }
    }
    int removeStones(vector<vector<int>>& stones) {
        for(auto i:stones){
            int id=i[0]*n+i[1];
            father[id]=id;
            x[i[0]].push_back(id);
            y[i[1]].push_back(id);
        }
        for(auto i:x){
            int id0=i.second[0];
            for(int j=1;j<i.second.size();j++){
                int id=i.second[j];
                if(find_father(id0)!=find_father(id)){
                    Union(id0,id);
                }
            }
        }
        for(auto i:y){
            int id0=i.second[0];
            for(int j=1;j<i.second.size();j++){
                int id=i.second[j];
                if(find_father(id0)!=find_father(id)){
                    Union(id0,id);
                }
            }
        }
        unordered_set<int> s;
        for(auto i:stones){
            int id=i[0]*n+i[1];
            s.insert(find_father(id));
        }
        return stones.size()-s.size();
    }
};
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