Leetcode counts the logarithm of points that cannot reach each other in an undirected graph

Give you an integer  n , Means a sheet of   Undirected graph   There is  n  Nodes , The number is  0  To  n - 1 . And give you a two-dimensional array of integers  edges , among  edges[i] = [ai, bi]  Representation node  ai  and  bi  There is a   Undirected   edge .

Please return   Unable to reach each other   Different   Number of pairs  .

Example 1：

 Input ：n = 3, edges = [[0,1],[0,2],[1,2]]
 Output ：0
 explain ： All points can reach each other , It means that you can't reach each other without point pairs , So we go back to  0 .

Example 2：

 Input ：n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
 Output ：14
 explain ： All in all  14  Point pairs cannot reach each other ：
[[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]]
 So we go back to  14 .

Tips ：

• 1 <= n <= 105
• 0 <= edges.length <= 2 * 105
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• No duplicate edges .

C++

class Solution {
public:
    int find(int x) {
        if(pre[x]==x) {
            return x;
        }
        int y=pre[x];
        int root=find(y);
        pre[x]=root;
        return root;
    }
    
    long long countPairs(int n, vector<vector<int>>& edges) {
        pre.resize(n);
        for(int i=0;i<n;i++) {
            pre[i]=i;
        }
        for(auto e:edges) {
            int x=e[0];
            int y=e[1];
            int root1=find(x);
            int root2=find(y);
            if(root1!=root2) {
                pre[root2]=root1;
            }
        }
        unordered_map<int,long> mp;
        for(int i=0;i<n;i++) {
            int root=find(i);
            mp[root]++;
        }
        long res=(long)n*(long)(n-1)/2;
        for(auto it:mp) {
            res-=it.second*(it.second-1)/2;
        }
        return res;
    }
private:
    vector<int> pre;
};