题意：给出n个点的坐标,要把n个点连通，使得总距离最小，可是有m对点已经连接，输入m，和m组a和b，表示a和b两点已经连接。

思路：两种做法。（1）用prim算法时，输入a，b。令mp[a][b]=0。然后进行一遍prim（2）Kruskal算法+并查集

代码：

//prim写法
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 2005;
const int MAXM = 200010;
const int N = 1005;

struct Node
{
    double x,y;
}node[maxn];

int n,m;
double mp[maxn][maxn];
double dist[maxn];
bool vis[maxn];

double Dis(Node n1,Node n2)
{
    return sqrt((n1.x-n2.x)*(n1.x-n2.x)+(n1.y-n2.y)*(n1.y-n2.y));
}

double prim()
{
    int i,j,now;
    double mi;
    mem(vis,false);
    mem(dist,INF);
    for (i=1;i<=n;i++)
        dist[i]=mp[1][i];
    dist[1]=0;
    vis[1]=true;
    for (i=1;i<=n;i++)
    {
        now=-1;
        mi=INF;
        for (j=1;j<=n;j++)
        {
            if (!vis[j]&&mi>dist[j])
            {
                now=j;
                mi=dist[j];
            }
        }
        if (now==-1) break;
        vis[now]=true;
        for (j=1;j<=n;j++)
        {
            if (!vis[j]&&dist[j]>mp[now][j])
                dist[j]=mp[now][j];
        }
    }
    double ans=0;
    for (i=1;i<=n;i++)
        ans+=dist[i];
    return ans;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v;
    while (~sf(n))
    {
        for (i=1;i<=n;i++)
            scanf("%lf%lf",&node[i].x,&node[i].y);
        for (i=1;i<=n;i++)
        {
            for (j=1;j<=n;j++)
                if (i==j) mp[i][j]=0;
                else mp[i][j]=Dis(node[i],node[j]);
        }
        sf(m);
        for (i=0;i<m;i++)   //在同一个联通块的距离直接赋为0
        {
            sff(u,v);
            mp[u][v]=mp[v][u]=0;
        }
        printf("%.2f\n",prim());
    }
    return 0;
}


//Kruskal+并查集写法
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 1000000;
const int MAXM = 200010;
const int N = 1005;

struct Node
{
    double x,y;
}node[maxn];

struct Edge
{
    int u,v;
    double len;
}edge[MAXN];

int n,m,cnt;
int father[maxn];

int cmp(Edge e1,Edge e2)
{
    return e1.len<e2.len;
}

void init()
{
    cnt=0;
    for (int i=0;i<=n;i++)
        father[i]=i;
}

double Dis(Node n1,Node n2)
{
    return sqrt((n1.x-n2.x)*(n1.x-n2.x)+(n1.y-n2.y)*(n1.y-n2.y));
}

int find_father(int x)
{
    if (x!=father[x])
        father[x]=find_father(father[x]);
    return father[x];
}

double Kruskal()
{
    int i,j;
    double ans=0;
    sort(edge,edge+cnt,cmp);
    for (i=0;i<cnt;i++)
    {
        int fu=find_father(edge[i].u);
        int fv=find_father(edge[i].v);
        if (fu!=fv)
        {
            ans+=edge[i].len;
            father[fu]=fv;
        }
    }
    return ans;
}


int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v;
    while (~sf(n))
    {
        init();
        for (i=1;i<=n;i++)
            scanf("%lf%lf",&node[i].x,&node[i].y);
        for (i=1;i<n;i++)
        {
            for (j=i+1;j<=n;j++)
            {
                if (i==j) continue;
                else
                {
                    double x=Dis(node[i],node[j]);
                    edge[cnt].u=i;
                    edge[cnt].v=j;
                    edge[cnt++].len=x;
                }
            }
        }
        sf(m);
        for (i=1;i<=m;i++)
        {
            sff(u,v);
            int fu=find_father(u);
            int fv=find_father(v);
            if (fu!=fv)
                father[fu]=fv;
        }
        printf("%.2f\n",Kruskal());
    }
    return 0;
}