通过1≤K≤17比较容易看出这道题目需要用状压dp,先处理出c[i]到其他点的最短距离，然后状压更新当前路径的最小答案，最后枚举一下终点得到答案。

// #pragma GCC optimize(2)
// #include <random>
// #include <windows.h>
// #include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <set>
#include <map>
#define IO                       \
    ios::sync_with_stdio(false); \
    // cout.tie(0);
#define lson(x) node << 1, start, mid
#define rson(x) node << 1 | 1, mid + 1, end
using namespace std;
// int dis[8][2] = {0, 1, 1, 0, 0, -1, -1, 0, 1, -1, 1, 1, -1, 1, -1, -1};
typedef unsigned long long ULL;
typedef long long LL;
typedef pair<int, int> P;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 10;
const int maxm = 1e2 + 10;
const LL mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1);
// int dis[4][2] = {1, 0, 0, -1, 0, 1, -1, 0};
// int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

struct Edge
{
    int next, to;
} e[maxn << 1];
int head[maxn];
int d[maxn];
int dis[20][20];
int f[1 << 19][20];
int a[20];
int n, m, kk;
int k = 0;

void add(int u, int v)
{
    e[k].next = head[u];
    e[k].to = v;
    head[u] = k++;
}
void BFS(int s)
{
    memset(d, inf, sizeof d);
    queue<int> q;
    q.push(s);
    d[s] = 0;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        for (int i = head[u]; ~i; i = e[i].next)
        {
            int v = e[i].to;
            if (d[v] == inf)
            {
                d[v] = d[u] + 1;
                q.push(v);
            }
        }
    }
    return;
}
int main()
{
#ifdef WXY
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    IO;
    int x, y;
    memset(head, -1, sizeof head);
    memset(dis, inf, sizeof dis);
    cin >> n >> m;
    for (int i = 0; i < m; i++)
    {
        cin >> x >> y;
        add(x, y);
        add(y, x);
    }
    cin >> kk;
    for (int i = 1; i <= kk; i++)
        cin >> a[i];
    for (int i = 1; i <= kk; i++)
    {
        BFS(a[i]);
        for (int j = 1; j <= kk; j++)
            dis[i][j] = d[a[j]];
    }
    memset(f, inf, sizeof f);

    for (int i = 0; i < kk; i++)
        f[1 << i][i + 1] = 1;

    int all = (1 << kk) - 1;
    for (int i = 1; i <= all; i++)
    {
        for (int j = 1; j <= kk; j++)
        {
            if (f[i][j] == inf)
                continue;
            for (int t = 1; t <= kk; t++)
            {
                if (j == t)
                    continue;
                if (!(i >> (t - 1) & 1))
                {
                    int nex = i + (1 << (t - 1));
                    f[nex][t] = min(f[nex][t], f[i][j] + dis[j][t]);
                }
            }
        }
    }
    int ans = inf;
    for (int i = 1; i <= kk; i++)
    {
        ans = min(ans, f[all][i]);
    }
    if (ans == inf)
        cout << -1;
    else
        cout << ans;
    return 0;
}