【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

  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 set<int>G[N],S;
 36 int vis[N],c[N],ans;
 37 
 38 int read()
 39 {
 40    int v=0,f=1;
 41    char c=getchar();
 42    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 43    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 44    return v*f;
 45 }
 46 
 47 ll readll()
 48 {
 49    ll v=0,f=1;
 50    char c=getchar();
 51    while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
 52    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
 53    return v*f;
 54 }
 55 
 56 void bfs(int st)
 57 {
 58     queue<int> q;
 59     S.erase(st);
 60     q.push(st);
 61     int s=1;
 62     while(q.size()>0)
 63     {
 64         int u=q.front();
 65         q.pop();
 66         if(vis[u]) continue;
 67         vis[u]=1;
 68 
 69         for(auto it=S.begin();it!=S.end();)
 70         {
 71             int v=*it;
 72             it++;
 73             if(G[u].find(v)==G[u].end())
 74             {
 75                 q.push(v);
 76                 S.erase(v);
 77                 s++;
 78             }
 79         }
 80     }
 81     c[ans]=s;
 82 }
 83 
 84 int main()
 85 {
 86     int n=read(),m=read();
 87     rep(i,1,n) S.insert(i);
 88     rep(i,1,m)
 89     {
 90         int x=read(),y=read();
 91         G[x].insert(y);
 92         G[y].insert(x);
 93     }
 94     ans=0;
 95     rep(i,1,n)
 96      if(!vis[i])
 97      {
 98          ans++;
 99          bfs(i);
100      }
101     printf("%d\n",ans);
102     sort(c+1,c+ans+1);
103     rep(i,1,ans) printf("%d ",c[i]);
104     return 0;
105 }

 

posted on 2019-11-07 15:34  myx12345  阅读(201)  评论(0)  编辑  收藏  举报

导航