Codeforces Round #600(Div.2) D 문항

Harmonious Graph
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 In other words, in a harmonious graph, if from a node l we can reach a node r through edges (l
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.
그리고 수집+사유;
각 집합의 최대 최소치 사이의 값을 매거하고 아버지가 같지 않으면 연결하고 변수에 1을 더한다.

#include
#define LL long long
#define pa pair
#define lson k<<1
#define rson k<<1|1
#define inf 0x3f3f3f3f
//ios::sync_with_stdio(false);
using namespace std;
const int N=200100;
const int M=200100;
const LL mod=998244353;
int n,m;
int fa[N];
int find(int p){
	if(p==fa[p]) return p;
	return fa[p]=find(fa[p]);
}
int vis[N];
struct Node{
	int mmin,mmax;
}bin[N];
bool cmp(Node p,Node q){
	return p.mmin<q.mmin;
}
bool d[N];
int main(){
//	ios::sync_with_stdio(false);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		bin[i].mmax=0;
		bin[i].mmin=2e9;
		fa[i]=i;
	}
	int u,v;
	int ans=0;//       
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		if(!vis[u]&&!vis[v]){
			ans++;
			vis[u]=vis[v]=ans;
		}
		else if(vis[u]||vis[v]){
			if(vis[u]) vis[v]=vis[u];
			else vis[u]=vis[v];
		}
		if(u>v){
			fa[find(v)]=find(fa[u]);
			bin[vis[u]].mmax=bin[vis[u]].mmax>u?bin[vis[u]].mmax:u;
			bin[vis[u]].mmin=bin[vis[u]].mmin<v?bin[vis[u]].mmin:v;
		}
		else{
			fa[find(u)]=find(fa[v]);
			bin[vis[u]].mmax=bin[vis[u]].mmax>v?bin[vis[u]].mmax:v;
			bin[vis[u]].mmin=bin[vis[u]].mmin<u?bin[vis[u]].mmin:u;
		}
	}
	sort(bin+1,bin+1+ans,cmp);
	int sum=0;
	int last=0;
	for(int i=1;i<=ans;i++){
		int f=find(bin[i].mmin);//        
		last=max(last,bin[i].mmin);
		for(int j=last;j<bin[i].mmax;j++){
			if(d[j]) continue;//   
			d[j]=true;
			if(find(fa[j])!=f){
				if(fa[j]==j){//         
					fa[j]=f;
					sum++;
				}
				else{
					if(fa[j]>f){//           
						fa[find(f)]=find(fa[j]);
						f=find(fa[j]);
					}
					else{
						fa[find(fa[j])]=find(f);
					}
					sum++;
				}
			}
		}
		last=bin[i].mmax;
	}
	printf("%d
",sum);
	return 0;
}