Codeforces Round 599 (Div 1 + 2.) ___ D 0-1 MST -. Optimization list BFS

Topic links: point I ah ╭ (╯ ^ ╰) ╮

Subject to the effect:

     $n$ complete graph points
    given $m$ edges, the right side of $1$， $0≤m≤min( \frac{n(n−1)}{2},10^5)$
    other side is right $0$
    minimum spanning tree

Problem-solving ideas:

    Title number of connected block diagram into complementing
    each enumeration one unprocessed point, a wide-search
    and enumerate the original on each side, marked with marker
    points and unlabeled certain point enumeration complement FIG belonging to a communication block
    these queued unlabeled point, a wide-search
    optimization processing chain with points, points on the treated delete list
    time complexity: $O(n+m)$

---

    The total complexity of enumeration side is: $Man)$
    enumerated points are divided into two cases:
    ①: the point is not marked, the queued each point only once enqueue
    ②: the point has been marked, the time complexity of the operator $Man)$ inner

Core: Pictured chain optimization complement dense map

#include<bits/stdc++.h>
#define rint register int
#define deb(x) cerr<<#x<<" = "<<(x)<<'\n';
using namespace std;
typedef long long ll;
using pii = pair <ll,int>;
const int maxn = 2e5 + 5;
int n, m, ans[maxn], cnt;
int l[maxn], r[maxn];
int vis[maxn], vis2[maxn];
vector <int> g[maxn];
 
inline void del(int x){
	l[r[x]] = l[x];
	r[l[x]] = r[x];
}
 
void bfs(int x){
	vis[x] = ans[++cnt] = 1;
	queue <int> Q;
	Q.push(x); del(x);
	while(Q.size()){
		int q = Q.front(); Q.pop();
		for(auto v : g[q])	
			if(!vis[v]) vis2[v] = 1;
		for(int i=r[0]; i; i=r[i])	
			if(!vis2[i]) {
				Q.push(i);
				del(i);
				vis[i] = 1;
				++ans[cnt];
			} else vis2[i] = 0;
	}
}
 
int main() {
	scanf("%d%d", &n, &m);
	for(int i=1, u, v; i<=m; i++) {
		scanf("%d%d", &u, &v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for(int i=0; i<=n; i++) l[i] = i-1, r[i] = i+1;
	r[n] = 0;
	for(int i=1; i<=n; i++)
		if(!vis[i]) bfs(i);
	printf("%d\n", cnt - 1);
}