题意：平面上有n个点(1<=n<=1000)，你的任务是让所有n个点连通。可以新建边，费用等于两端点欧几里德距离的平方。也可以购买套餐（套餐中的点全部连通）。问最小费用。
分析：
1、先将不购买任何套餐的最小生成树的所有边(边数为cnt)存起来，目的是枚举套餐时不必再耗Kruskal算法的O(n2)复杂度，而是降低为O(cnt)。
2、二进制枚举套餐。
3、枚举套餐时，先将套餐中的边按最小生成树建边，在将不购买任何套餐的最小生成树的cnt条边建上，因为套餐中的边权值为0，所以这样处理不会影响结果。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
vector<int> v;//不选择套餐时最小生成树的边编号
int ans, n, m;
int fa[MAXN];
struct City{//城市
int x, y, id;
void read(){
scanf("%d%d", &x, &y);
}
}num[MAXN];
struct Package{//套餐
int n, cost;
int city[MAXN];
void read(){
scanf("%d%d", &n, &cost);
for(int i = 0; i < n; ++i){
scanf("%d", &city[i]);
}
}
}q[10];
struct Edge{
int from, to, dist;
void set(int f, int t, int d){
from = f;
to = t;
dist = d;
}
bool operator < (const Edge& rhs)const{
return dist < rhs.dist;
}
}e[MAXN * MAXN];
int getD(City& A, City& B){
return (A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y);
}
int Find(int v){
return fa[v] = (fa[v] == v) ? v : Find(fa[v]);
}
void solve(){//二进制枚举子集
for(int i = 1; i < (1 << m); ++i){
int tmp = 0;
for(int j = 1; j <= n; ++j) fa[j] = j;//初始化并查集
for(int j = 0; j < m; ++j){
if(i & (1 << j)){
tmp += q[j].cost;
for(int a = 0; a < q[j].n; ++a){
for(int b = a + 1; b < q[j].n; ++b){
int x = Find(q[j].city[a]);
int y = Find(q[j].city[b]);
if(x == y) continue;
if(x < y) fa[y] = x;
else fa[x] = y;
}
}
}
}
int len = v.size();
for(int j = 0; j < len; ++j){
int id = v[j];
int x = Find(e[id].from);
int y = Find(e[id].to);
if(x == y) continue;
tmp += e[id].dist;
if(x < y) fa[y] = x;
else fa[x] = y;
}
ans = Min(ans, tmp);
}
}
int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%d%d", &n, &m);
for(int i = 0; i < m; ++i){
q[i].read();
}
for(int i = 1; i <= n; ++i){
num[i].read();
fa[i] = i;
}
int cnt = 0;
for(int i = 1; i <= n; ++i){
for(int j = i + 1; j <= n; ++j){
e[cnt++].set(i, j, getD(num[i], num[j]));
}
}
sort(e, e + cnt);
v.clear();
ans = 0;
for(int i = 0; i < cnt; ++i){
int x = Find(e[i].from);
int y = Find(e[i].to);
if(x == y) continue;
ans += e[i].dist;
v.push_back(i);
if(x < y) fa[y] = x;
else fa[x] = y;
}
solve();
printf("%d\n", ans);
if(T) printf("\n");
}
return 0;
}