题意：

n≤2e5,m≤2e5n\le2e5,m\le2e5n≤2e5,m≤2e5。

思路：

这是题是我上个题的一部分，算是个小知识点，暴力能过。
直接维护一个setsetset，让后遍历所有点，如果当前这个点没有被遍历，那么就放入队列中，扩展出去，每个点最多入队一次，而且边数不会很多，复杂度大约是(n+m)logn(n+m)logn(n+m)logn

// Problem: E. Connected Components?
// Contest: Codeforces - Educational Codeforces Round 37 (Rated for Div. 2)
// URL: https://codeforces.com/problemset/problem/920/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m;
int p[N],se[N];
LL rest;
bool st[N];
set<int>v[N],all;
struct Node {int a,b,w,flag; bool operator < (const Node &W) const {return w<W.w;}
}edge[N];int find(int x) {return x==p[x]? x:p[x]=find(p[x]);
}void get_block() {queue<int>q;vector<int>now;for(int i=1;i<=n;i++) {if(!st[i]) {q.push(i); st[i]=1;all.erase(i);while(q.size()) {int u=q.front(); q.pop();now.clear();for(auto x:all) {if(v[u].count(x)) continue;q.push(x); now.pb(x);st[x]=1;int a=find(u),b=find(x);se[a]+=se[b];p[b]=a;rest--;}for(auto x:now) all.erase(x);}}}
}int main()
{//  ios::sync_with_stdio(false);
//  cin.tie(0);cin>>n>>m;rest=1ll*n*(n-1)/2-m;for(int i=1;i<=n;i++) all.insert(i),p[i]=i;for(int i=1;i<=m;i++) {int a,b; scanf("%d%d",&a,&b);v[a].insert(b); v[b].insert(a);}for(int i=1;i<=n;i++) p[i]=i,se[i]=1;get_block();vector<int>ans;for(int i=1;i<=n;i++) if(i==find(i)) ans.pb(se[i]);sort(ans.begin(),ans.end());printf("%d\n",ans.size());for(auto x:ans) printf("%d ",x);return 0;
}
/**/

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Educational Codeforces Round 37 (Rated for Div. 2) E. Connected Components? 暴力 + 补图的遍历

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题意：

思路：

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