I found that hard to deal with edging operation, so consider the first increase after completion of the border erase all sides.
Deleting pair of side x to the y- if both have an apparently skipped school does not make sense, so if you have skipped school so for them to judge, if you can not cut class to go to continue the search.
This appears to be a time complexity o (nm) , but notes that each point of up to once cut class becomes not cut class, so is o (n + m) is.
1 #include<bits/stdc++.h> 2 using namespace std; 3 queue<int>q; 4 struct ji{ 5 int nex,to; 6 }edge[400001]; 7 int E,n,m,k,head[200001],x[200001],y[200001],d[200001],ans[200001]; 8 bool vis[200001],flag[400001]; 9 void add(int x,int y){ 10 edge[E].nex=head[x]; 11 edge[E].to=y; 12 head[x]=E++; 13 } 14 void pus(int t){ 15 if (d[t]<k){ 16 q.push(t); 17 ans[0]-=(vis[t]=1); 18 } 19 } 20 void erase(){ 21 while (!q.empty()){ 22 int k=q.front(); 23 q.pop(); 24 for(int i=head[k];i!=-1;i=edge[i].nex) 25 if ((!vis[edge[i].to])&&(!flag[i])){ 26 flag[i]=flag[i^1]=1; 27 d[edge[i].to]--; 28 pus(edge[i].to); 29 } 30 } 31 } 32 int main(){ 33 memset(head,-1,sizeof(head)); 34 scanf("%d%d%d",&n,&m,&k); 35 for(int i=1;i<=m;i++){ 36 scanf("%d%d",&x[i],&y[i]); 37 add(x[i],y[i]); 38 add(y[i],x[i]); 39 d[x[i]]++; 40 d[y[i]]++; 41 } 42 ans[0]=n; 43 for(int i=1;i<=n;i++)pus(i); 44 for(int i=m;i;i--){ 45 erase(); 46 ans[i]=ans[0]; 47 if (vis[x[i]]+vis[y[i]]==0){ 48 d[x[i]]--; 49 d[y[i]]--; 50 } 51 flag[i*2-1]=flag[i*2-2]=1; 52 if (!vis[x[i]])pus(x[i]); 53 if (!vis[y[i]])pus(y[i]); 54 } 55 for(int i=1;i<=m;i++)printf("%d\n",ans[i]); 56 }