Buy or Build UVA

题意：

大概意思是有 n 个点，现在有 q 个方案 ，第 i 个方案耗费为 ci ，使 Ni 个点联通 ，当然也可以直接使两点联通 ，现求最小生成树的代价。

两点直接联通的代价是欧几里得距离的平方；

 

由于0<=q<=8，所以我们考虑二进制枚举；

该位为1表示选择该方案，然后每次求一遍cost ，最后取 min 即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}
int T;
int n, q;
int cost[maxn];

int fa[maxn];
int cnt;
vector<int>vc[maxn];

struct point {
	int x, y;
}pint[maxn];

struct node {
	int x, y;
	int w;
}edge[maxn];

bool cmp(node a, node b) {
	return a.w < b.w;
}

int dis(int a, int b, int x, int y) {
	return ((a - x)*(a - x) + (b - y)*(b - y));
}

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

int findfa(int x) {
	if (x == fa[x])return x;
	return fa[x] = findfa(fa[x]);
}

void Union(int p, int q) {
	if (findfa(q) != findfa(p)) {
		fa[findfa(p)] = findfa(q);
	}
}

int ans;

int kruskal() {
	int res = 0;
	int ct = 0;
	for (int i = 0; i < cnt; i++) {
		int u = edge[i].x; int v = edge[i].y;
		if (findfa(u) != findfa(v)) {
			Union(u, v); res += edge[i].w;
			ct++;
			if (ct == n - 1)break;
		}
	}
	return res;
}


void sol() {
	init(n);
	ans = kruskal();
	
	for (int i = 0; i < (1 << q); i++) {
		init(n);int cst = 0;
		for (int j = 0; j < q; j++) {
			if (((i >> j) & 1)==0)continue;
			cst += cost[j];
			for (int k = 1; k < vc[j].size(); k++) {
				Union(vc[j][k], vc[j][0]);
			}
		}
		ans = min(ans, cst + kruskal());
	}
	printf("%d
", ans);
}

int main()
{
	//ios::sync_with_stdio(0);
	rdint(T); int kase = 1;
	while (T--) {
		if (kase > 1)printf("
");
		kase++;
		rdint(n); rdint(q); cnt = 0;
		for (int i = 0; i < 10; i++)vc[i].clear();
		for (int i = 0; i < q; i++) {
			int tmp; rdint(tmp); rdint(cost[i]);
			while (tmp--) {
				int x; rdint(x); vc[i].push_back(x);
			}
		}
		
		for (int i = 1; i <= n; i++) {
			rdint(pint[i].x); rdint(pint[i].y);
		}
		for (int i = 1; i <= n; i++) {
			for (int j = i + 1; j <= n; j++) {
				edge[cnt].x = i; edge[cnt].y = j; edge[cnt].w = dis(pint[i].x, pint[i].y, pint[j].x, pint[j].y); cnt++;
			}
		}
		sort(edge, edge + cnt, cmp);
		sol();
		
	}

    return 0;
}