bfs+状态压缩求出所有的状态，然后由于第一个节点需要特殊处理，可以右移一位剔除掉，也可以特判。然后采用集合的操作，

#pragma comment(linker,"/STACK:1024000000,1024000000")

#include <cstdio>

#include <queue>

using namespace std;

#define inf 0x3f3f3f3f

int n, m, cnt;

int head[17], next[17 * 17 * 2 + 3][3], dp[1 << 17][17], dis[1 << 17], num[1 << 17];

void add (int u, int v, int w)

{

	next[cnt][1] = v;

	next[cnt][2] = w;

	next[cnt][0] = head[u];

	head[u] = cnt++;

}

void bfs ()

{

	queue<pair<int, int> > q;

	q.push(make_pair(0, 1));

	dp[1][0] = 0;

	while (!q.empty()){

		pair<int, int> p = q.front();

		q.pop();

		int su = p.second;

		int u = p.first;

		for (int i = head[u]; i != -1; i = next[i][0]){

			int v = next[i][1];

			int w = next[i][2];

			int sv = su|(1 << v);

			if(dp[sv][v] > dp[su][u] + w){

				dp[sv][v] = dp[su][u] + w;

				q.push(make_pair(v, sv));

			}

		}

	}

}

int main ()

{

	//freopen ("in.txt", "r", stdin);

	int t, count = 0;

	scanf ("%d", &t);

	while (t--)

	{

		scanf ("%d %d", &n, &m);

		int u, v, w;

		for (int i = 0; i < n; ++i) head[i] = -1;

		for (int i = (1 << n) - 1; i >= 0; --i){

			dis[i] = inf;

			num[i] = inf;

			for (int j = 0; j < n; ++j) dp[i][j] = inf;

		}

		cnt = 0;

		for (int i = 0; i < m; ++i){

			scanf ("%d %d %d", &u, &v, &w);

			add (u - 1, v - 1, w);

			add (v - 1, u - 1, w);

		}

		scanf ("%d", &m);

		v = 0;

		for (int i = 0; i < m; ++i){

			scanf ("%d", &u);

			v |= (1 << (u - 1));

		}

		v >>= 1;

		if (!m || (m == 1 && u == 1)){

			printf("Case %d: 0\n", ++count);

			continue;

		}

		bfs ();

		u = inf;

		w = (1 << (n-1)) - 1;

		for(int i = 1; i < (1 << n); ++i)

			for(int j = 0; j < n; ++j)

				dis[i >> 1] = min(dis[i >> 1], dp[i][j]);

		for (int i = 1; i <= w; ++i)

			for (int j = i; j; j = (j - 1) & i)

				num[i] = min(num[i], max(dis[j], dis[i ^ j]));

		for (int i = 1; i <= w; ++i)

			for(int j = i; j; j = (j - 1) & i)

			if((i & v) == v) u = min(u, max(num[j], dis[i ^ j]));

		if (u == inf) u = -1;

		printf("Case %d: %d\n", ++count, u);

	}

	return 0;

}