【BZOJ1098】办公楼biu（补图，bfs，链表）

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

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef long double ld;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 typedef pair<ll,ll>P;
 #define N  500010
 #define M  1000000
 #define INF 1e9
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
 #define fors(i) for(auto i:e[x]) if(i!=p)
 
 const int MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int dx[]={-,,,};
       int dy[]={,,-,};
 
 VI c[N];
 int pre[N],nxt[N],vis[N],q[N],a[N],ans,n,m;
 
 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
 
 ll readll()
 {
    ll v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
 
 void del(int u)
 {
     int t=pre[u];
     nxt[t]=nxt[u];
     pre[nxt[u]]=t;
 }
 
 void bfs(int st)
 {
     int t=,w=;
     q[]=st;
     while(t<w)
     {
         a[ans]++;
         t++;
         int u=q[t];
         for(int i=;i<c[u].size();i++)
         {
             int v=c[u][i];
             vis[v]=;
         }
         for(int i=nxt[];i<=n;i=nxt[i])
          if(!vis[i]) del(i),q[++w]=i;
         for(int i=;i<c[u].size();i++)
         {
             int v=c[u][i];
             vis[v]=;
         }
     }
 }
 
 int main()
 {
     n=read(),m=read();
     rep(i,,n) nxt[i]=i+;
     rep(i,,n+) pre[i]=i-;
     rep(i,,m)
     {
         int x=read(),y=read();
         c[x].pb(y);
         c[y].pb(x);
     }
     ans=;
     for(int i=nxt[];i<=n;i=nxt[])
     {
         del(i);
         ans++;
         bfs(i);
     }
     printf("%d\n",ans);
     sort(a+,a+ans+);
     rep(i,,ans) printf("%d ",a[i]);
     return ;
 }