D. Harmonious Graph (Codeforces Round # 600 (Div. 2)) (disjoint-set thinking +)

D. Harmonious Graph (Codeforces Round # 600 (Div. 2)) (disjoint-set thinking +)

input

14 8
1 2
2 7
3 4
6 3
5 7
3 8
6 8
11 12

output

1

input

200000 3
7 9
9 8
4 5

output

0

Example 1:

answer

If there is $1-n$ path, then $1-n$ The communication block must be able to reach $2,3,4...11$ all points. And if and $1$ maximum point of attachment is $7$ So, $1-7$ communicates and blocks only $2,3,4,5,6$ connected on the line. Therefore, we maintain $lar$ maximum value for all blocks in the current through the communication of, just need to find less than $lar$ of this point to explain that there is an edge here, because the title already gives some edge, so that part is minus effective among these answers.

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 200020;

int res;
int lab[N];
int n, m;

int find(int x) {
	return lab[x] < 0 ? x : lab[x] = find(lab[x]);
}

void join(int x, int y) {
	x = find(x); y = find(y);
	if (x == y) return;
	if (x < y) swap(x, y);
	res++;
	lab[y] = x;
}

int main() {
	//freopen("in.txt", "r", stdin);
	ios_base::sync_with_stdio(false); cin.tie(NULL);

	memset(lab, -1, sizeof lab);

	cin >> n >> m;
	res = 0;
	for (int i = 1; i <= m; ++i) {
		int u, v; cin >> u >> v;
		join(u, v);
	}

	int comp = 0, lar = find(1);
	for (int i = 2; i <= n; ++i) {
		if (i <= lar) ++comp;
		lar = max(lar, find(i));
	}

	cout << comp - res << endl;

	return 0;
}