D. Harmonious Graph

Description:
You're given an undirected graph with n nodes and m edges. Nodes are numbered from 1 to n.

The graph is considered harmonious if and only if the following property holds:

For every triple of integers (l,m,r) such that 1≤l<m<r≤n, if there exists a path going from node l to node r, then there exists a path going from node l to node m.
In other words, in a harmonious graph, if from a node l we can reach a node r through edges (l<r), then we should able to reach nodes (l+1),(l+2),…,(r−1) too.

What is the minimum number of edges we need to add to make the graph harmonious?

Input
The first line contains two integers n and m (3≤n≤200 000 and 1≤m≤200 000).

The i-th of the next m lines contains two integers ui and vi (1≤ui,vi≤n, ui≠vi), that mean that there's an edge between nodes u and v.

It is guaranteed that the given graph is simple (there is no self-loop, and there is at most one edge between every pair of nodes).

Output
Print the minimum number of edges we have to add to the graph to make it harmonious.

Examples
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
Note
In the first example, the given graph is not harmonious (for instance, 1<6<7, node 1 can reach node 7 through the path 1→2→7, but node 1 can't reach node 6). However adding the edge (2,4) is sufficient to make it harmonious.

In the second example, the given graph is already harmonious.

Problem solving:

Code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int       f[maxn];
int find(int x)
{
return f[x] != x ? f[x] = find(f[x]) : x;
}
void join(int x, int y)
{
x = find(x), y = find(y);
if (x != y)
{
if (x < y)
swap(x, y);
f[x] = y;
}
}
int main()
{
ios::sync_with_stdio(0);
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++)
f[i] = i;
for (int i = 0, u, v; i < m; i++)
{
cin >> u >> v;
join(u, v);
}
int ans = 0, mid = find(n);
for (int i = n; i >= 1; i--)
{
if (i < mid)
mid = find(i);
else if (find(i) != mid)
join(i, mid), ans++;
mid = min(find(i), mid);
}
cout << ans << endl;
return 0;
}