1 #include<iostream>
  2 #include<algorithm>
  3 #include<cmath>
  4 using namespace std;
  5
  6 const int maxn = 100+10;
  7 int far[maxn];
  8 int n, m;
  9 int k;
 10 struct Edge
 11 {
 12     const bool operator<(Edge &n)const
 13     {
 14         return value<n.value;
 15     }
 16     int start;
 17     int end;
 18     double value;
 19 }edge[maxn*maxn];
 20
 21 bool cmp(struct Edge &a, struct Edge &b)
 22 {
 23     return a.value<b.value;
 24 }
 25
 26 void init()
 27 {
 28     int i, j;
 29     for(i=0; i<=n; i++)
 30         far[i] = i;
 31 }
 32
 33 int find(int a)
 34 {
 35     if(far[a]!=a)
 36         far[a] = find(far[a]);
 37     return far[a];
 38 }
 39
 40 void Union(int a, int b)
 41 {
 42     a = find(a);
 43     b = find(b);
 44     if(a != b)
 45         far[a] = b;
 46 }
 47
 48 double kruskal()
 49 {
 50     init();
 51     sort(edge,edge+n*(n-1)+1,cmp);
 52     double sum = 0;
 53     int count = 0;
 54     for(int i=1; i<=n*(n-1) && count<n; i++)
 55     {
 56         int a=edge[i].start;
 57         int b=edge[i].end;
 58         if(find(a)!=find(b))
 59         {
 60             Union(a, b);
 61             sum+=edge[i].value;
 62             count++;
 63         }
 64     }
 65     return sum;
 66 }
 67
 68 int main()
 69 {
 70     double a[maxn], b[maxn];
 71     while(cin>>n)
 72     {
 73         //m = n*(n-1)/2;
 74
 75         for(int i=1; i<=n; i++)
 76         {
 77             scanf("%lf%lf",&a[i], &b[i]);  //因为是无向图，所以要两次赋值
 78         }
 79         k=1;
 80         double c;
 81         for(int i=1; i<n; i++)
 82         {
 83             for(int j=i+1; j<=n; j++)
 84             {
 85                 //if(i != j)
 86                 //{ 87                     c = sqrt((a[i]-a[j])*(a[i]-a[j])+(b[i]-b[j])*(b[i]-b[j]));
 88                     edge[k].start = i;
 89                     edge[k].end = j;
 90                     edge[k++].value = c;
 91                     edge[k].start = j;
 92                     edge[k].end = i;
 93                     edge[k++].value = c;
 94                 //}
 95             }
 96         }
 97         printf("%.2lf\n", kruskal());
 98     }
 99     return 0;
100 }