【CF1243D&CF920E】0-1 MST（bfs，set）

题意：给定一张n个点的完全图，其中有m条边权为1其余为0，求最小生成树的权值和

n，m<=1e5

思路：答案即为边权为0的边连接的联通块个数-1

用set存图和一个未被选取的点的集合，bfs过程中如果找到边权为0且未被选取的边则加入

如果要维护联通块大小也在bfs里随便记一下就好

具体实现看代码

 #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[]={,,-,};

 set<int>G[N],S;
 int vis[N],c[N],ans;

 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 bfs(int st)
 {
     queue<int> q;
     S.erase(st);
     q.push(st);
     int s=;
     while(q.size()>)
     {
         int u=q.front();
         q.pop();
         if(vis[u]) continue;
         vis[u]=;

         for(auto it=S.begin();it!=S.end();)
         {
             int v=*it;
             it++;
             if(G[u].find(v)==G[u].end())
             {
                 q.push(v);
                 S.erase(v);
                 s++;
             }
         }
     }
     c[ans]=s;
 }

 int main()
 {
     int n=read(),m=read();
     rep(i,,n) S.insert(i);
     rep(i,,m)
     {
         int x=read(),y=read();
         G[x].insert(y);
         G[y].insert(x);
     }
     ans=;
     rep(i,,n)
      if(!vis[i])
      {
          ans++;
          bfs(i);
      }
     printf("%d\n",ans);
     sort(c+,c+ans+);
     rep(i,,ans) printf("%d ",c[i]);
     return ;
 }