Portal

I really don't know what algorithm is used.

Because the second qqq maximum independent set is npcnpcnpc problem
Consider maximizing ppp first
This can be maintained directly with the heap to minimize the current degree.

Then consider such a violent approach.
Each time the minimum point of the current degree is taken, and all adjacent points are deleted.

Because it is impossible to get a point whose degree is greater than ppp (otherwise the ppp of the current subgraph is larger than that of the first question).
So delete at most ppp points at a time
The total number of points will be N P + 1 lceil frac n {p + 1} rceil p+1n.
So it must have met the conditions.

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>  
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=10005;
vector<int> e[N];
int n,m,p,q;
int in[N],du[N],vis[N],del[N],stk[N],top;
priority_queue<pii,vector<pii>,greater<pii> > s;
char obuf[RLEN],*olen=obuf;
#define pc(x) (*(olen++)=(x))
inline void write(int x){
	(x>=10)&&(write(x/10),1);
	pc(x%10+'0');
}
inline void solve(){
	n=read(),m=read();
	for(int i=1;i<=m;i++){
		int u=read(),v=read();
		in[u]++,in[v]++,e[u].pb(v),e[v].pb(u);
	}
	for(int i=1;i<=n;i++)du[i]=in[i],s.push(pii(in[i],i)),vis[i]=0;
	while(!s.empty()){
		int u=s.top().se;s.pop();
		if(vis[u])continue;
		p=du[u],vis[u]=1;top=0;
		for(int &v:e[u])if(!vis[v]){
			du[v]--,s.push(pii(du[v],v));
		}
		stk[++top]=u;
		while(!s.empty()&&s.top().fi<=p){
			int u=s.top().se;s.pop();
			if(vis[u])continue;
			vis[u]=1;stk[++top]=u;
			for(int &v:e[u])if(!vis[v]){
				du[v]--,s.push(pii(du[v],v));
			}
		}
	}
	write(top);pc(' ');while(s.size())s.pop();
	for(int i=1;i<=top;i++)write(stk[i]),pc(' ');pc('\n');
	top=0;q=n/(p+1);
	for(int i=1;i<=n;i++)du[i]=in[i],s.push(pii(in[i],i)),vis[i]=0,del[i]=0;
	while(!s.empty()){
		int u=s.top().se;s.pop();
		if(vis[u]||del[u])continue;
		vis[u]=1;stk[++top]=u;if(top>=q)break;
		for(int &v:e[u]){
			del[v]=1;
			for(int &x:e[v])if(!del[x]&&!vis[x]){
				du[x]--,s.push(pii(du[x],x));
			}
		}
	}
	write(top),pc(' '); 
	for(int i=1;i<=top;i++)write(stk[i]),pc(' ');pc('\n');
	for(int i=1;i<=n;i++)e[i].clear(),in[i]=0;
	top=0;while(s.size())s.pop();
}
int main(){
	int T=read();
	while(T--)solve();
	fwrite(obuf,1,olen-obuf,stdout);
}