题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1626

题意：略。

思路：最小生成树变形，我们把已经建好道路的边权设为0然后跑下最小生成树就可以了。

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define MOD 1000000007
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define Key_Value ch[ch[root][1]][0]
#define DBN1(a)           cerr<<#a<<"="<<(a)<<"\n"
#define DBN2(a,b)         cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<"\n"
#define DBN3(a,b,c)       cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<"\n"
#define DBN4(a,b,c,d)     cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<"\n"
#define DBN5(a,b,c,d,e)   cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<"\n"
#define DBN6(a,b,c,d,e,f) cerr<<#a<<"="<<(a)<<", "<<#b<<"="<<(b)<<", "<<#c<<"="<<(c)<<", "<<#d<<"="<<(d)<<", "<<#e<<"="<<(e)<<", "<<#f<<"="<<(f)<<"\n"
#define clr(a,x) memset(a,x,sizeof(a))
using namespace std;
typedef long long ll;
const int maxn=1000+5;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
template<typename T> inline T gcd(T&a,T&b){return b==0?a:gcd(b,a%b);}
template<typename T> inline T lcm(T&a,T&b){return a/gcd(a,b)*b;}
template<typename T>
inline T read(T&x){
    x=0;int _f=0;char ch=getchar();
    while(ch<'0'||ch>'9')_f|=(ch=='-'),ch=getchar();
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x=_f?-x:x;
}
struct Edge{
    int u,v;
    double len;
    Edge(int _,int __,double ___):u(_),v(__),len(___){}
    friend bool operator<(const Edge&a,const Edge&b){
        return a.len<b.len;
    }
};
int n,m;
vector<Edge>edges;
int fa[maxn];
double x[maxn],y[maxn];
double cnt(int a,int b){return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));}
int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
int main(){
    read(n),read(m);
    for (int i=1;i<=n;i++) scanf("%lf%lf",&x[i],&y[i]),fa[i]=i;
    for (int i=1;i<=n;i++){
        for (int j=1;j<i;j++){
            double len=cnt(i,j);
            edges.push_back(Edge(i,j,len));
        }
    }
    for (int i=1;i<=m;i++){
        int u,v;read(u),read(v);
        edges.push_back(Edge(u,v,0));
    }
    sort(edges.begin(),edges.end());
    int c=0;
    double ans=0;
    for (int i=0;i<(int)edges.size();i++){
        Edge&e=edges[i];
        int u=e.u,v=e.v;
        int tx=Find(u),ty=Find(v);
        if (tx!=ty){
            fa[tx]=ty;
            ans+=e.len;
            if (++c>=n-1) break;
        }
    }
    printf("%.2lf\n",ans);
    return 0;
}