1 /* 2 用kuskal会被卡 3 一边prim一边算两点间距离（边权） 4 否则会MLE 5 另：不开long long见祖宗 6 */ 7 #include<bits/stdc++.h> 8 using namespace std; 9 const int maxn=5005; 10 const int INF=0x3f3f3f3f; 11 typedef pair<double,int> P; 12 priority_queue< P,vector<P>,greater<P> >q; 13 bool v[maxn]; 14 int n,m; 15 double ans,dis[maxn]; 16 struct point{int x,y;}d[maxn]; 17 double Dis(const point &x,const point &y){return sqrt((long long)(x.x-y.x)*(x.x-y.x)+(long long)(x.y-y.y)*(x.y-y.y));} 18 void prim() 19 { 20 fill(dis,dis+1+n,INF);dis[1]=0; 21 q.push(make_pair(0,1)); 22 while(!q.empty()) 23 { 24 int Top=q.top().second;q.pop(); 25 if(v[Top]) continue;ans+=dis[Top];v[Top]=1; 26 for(int i=1;i<=n;++i) 27 { 28 if(!v[i]) 29 { 30 double tmp=Dis(d[i],d[Top]); 31 if(tmp<dis[i]) 32 { 33 dis[i]=tmp; 34 q.push(make_pair(dis[i],i)); 35 } 36 } 37 } 38 } 39 } 40 int main() 41 { 42 scanf("%d",&n); 43 for(int i=1;i<=n;++i) scanf("%d%d",&d[i].x,&d[i].y); 44 prim(); 45 printf("%.2lf",ans); 46 return 0; 47 }