leetcode

0 class UnionFind {

1 int n, cnt = n;

2 vector<int> root, size;

3

4 public:

5 UnionFind(int n) : n(n), root(n), size(n, 1) { iota(root.begin(), root.end(), 0); }

6

7 int find(int x) {

8 while (x != root[x])

9 x = root[x] = root[root[x]];

10 return x;

11 }

12

13 void join(int x, int y) {

14 x = find(x), y = find(y);

15 if (x != y) {

16 if (size[x] > size[y]) swap(x, y);

17 root[x] = y;

18 size[y] += size[x];

19 cnt--;

20 }

21 }

22

23 int count() { return cnt; }

24 bool connected(int x, int y) { return find(x) == find(y); }

25

26 long long count_unreachable() {

27 long long res = 0;

28

29 for (int i = 0; i < n; i++)

30 if (root[i] == i) res += (long long)size[i] * (n - size[i]);

31

32 return res / 2;

33 }

34 };

35

36 class Solution {

37 public:

38 long long countPairs(int n, vector<vector<int>> &edges) {

39 UnionFind uf(n);

40 for (auto &e : edges)

41 uf.join(e[0], e[1]);

42 return uf.count_unreachable();

43 }

44 };