题意：有n个点m条边，要求将点尽可能多的分成若干个部分，使得若两个点不在同一个部分则他们之间必定有边
n<=1e5，m<=2e6
思路：From https://blog.csdn.net/clover_hxy/article/details/52980373
转化为求补图联通分量的个数和size，主要是枚举点用链表优化，整体可以做到不带log
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef long double ld; 7 typedef pair<int,int> PII; 8 typedef pair<ll,ll> Pll; 9 typedef vector<int> VI; 10 typedef vector<PII> VII; 11 typedef pair<ll,ll>P; 12 #define N 500010 13 #define M 1000000 14 #define INF 1e9 15 #define fi first 16 #define se second 17 #define MP make_pair 18 #define pb push_back 19 #define pi acos(-1) 20 #define mem(a,b) memset(a,b,sizeof(a)) 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 23 #define lowbit(x) x&(-x) 24 #define Rand (rand()*(1<<16)+rand()) 25 #define id(x) ((x)<=B?(x):m-n/(x)+1) 26 #define ls p<<1 27 #define rs p<<1|1 28 #define fors(i) for(auto i:e[x]) if(i!=p) 29 30 const int MOD=1e9+7,inv2=(MOD+1)/2; 31 double eps=1e-6; 32 int dx[4]={-1,1,0,0}; 33 int dy[4]={0,0,-1,1}; 34 35 VI c[N]; 36 int pre[N],nxt[N],vis[N],q[N],a[N],ans,n,m; 37 38 39 int read() 40 { 41 int v=0,f=1; 42 char c=getchar(); 43 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 44 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 45 return v*f; 46 } 47 48 ll readll() 49 { 50 ll v=0,f=1; 51 char c=getchar(); 52 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 53 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 54 return v*f; 55 } 56 57 void del(int u) 58 { 59 int t=pre[u]; 60 nxt[t]=nxt[u]; 61 pre[nxt[u]]=t; 62 } 63 64 void bfs(int st) 65 { 66 int t=0,w=1; 67 q[1]=st; 68 while(t<w) 69 { 70 a[ans]++; 71 t++; 72 int u=q[t]; 73 for(int i=0;i<c[u].size();i++) 74 { 75 int v=c[u][i]; 76 vis[v]=1; 77 } 78 for(int i=nxt[0];i<=n;i=nxt[i]) 79 if(!vis[i]) del(i),q[++w]=i; 80 for(int i=0;i<c[u].size();i++) 81 { 82 int v=c[u][i]; 83 vis[v]=0; 84 } 85 } 86 } 87 88 int main() 89 { 90 n=read(),m=read(); 91 rep(i,0,n) nxt[i]=i+1; 92 rep(i,1,n+1) pre[i]=i-1; 93 rep(i,1,m) 94 { 95 int x=read(),y=read(); 96 c[x].pb(y); 97 c[y].pb(x); 98 } 99 ans=0; 100 for(int i=nxt[0];i<=n;i=nxt[0]) 101 { 102 del(i); 103 ans++; 104 bfs(i); 105 } 106 printf("%d ",ans); 107 sort(a+1,a+ans+1); 108 rep(i,1,ans) printf("%d ",a[i]); 109 return 0; 110 }