這道題有很多解法，我是按照題解的做法寫了一遍。

題意：給出一個有向圖，然後有k個點，問這k個點兩兩之間的最短距離的最小值是多少。

解法：題解是把k個點按照每個點二進位制位的不同分成兩個集合，然後求一次多源最短路。由於最大的數是1e5，所以也就到個。。。2^17的樣子？由於是有向圖，還需要正反都跑一遍。所以一共跑34次最短路就出來了。。。

程式碼如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<stack>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
typedef pair<long long, int> pii;
const int maxn = 1e5 + 5;
const int maxm = 2e5 + 5;
const long long INF = 1e18;
int a[maxn];
int head[maxn], to[maxm], nx[maxm], cost[maxm], ppp = 0;

void add_edge(int u, int v, int val) {
    to[ppp] = v, nx[ppp] = head[u], cost[ppp] = val, head[u] = ppp++;
}

priority_queue <pii, vector<pii>, greater<pii> > q;
long long dis[maxn];

void dijkstra() {
    while(!q.empty()) {
        pii tmp = q.top();
        q.pop();
        int u = tmp.second;
        if(dis[u] < tmp.first)
            continue;
        for(int i = head[u]; ~i; i = nx[i]) {
            int v = to[i];
            if(dis[v] > dis[u] + cost[i]) {
                dis[v] = dis[u] + cost[i];
                q.push(pii(dis[v], v));
            }
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif
    int T, Case = 1;
    scanf("%d", &T);
    while(T--) {
        memset(head, -1, sizeof(head));
        ppp = 0;
        int n, m;
        scanf("%d%d", &n, &m);
        while(m--) {
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            add_edge(u, v, c);
        }
        int k;
        scanf("%d", &k);
        for(int i = 0; i < k; i++) {
            scanf("%d", &a[i]);
        }
        int ind = 1;
        long long Min = 1e18;
        while(1) {
            int cnt = 0;
            fill(dis, dis + n + 1, INF);
            for(int i = 0; i < k; i++) {
                if(a[i] & ind) {
                    dis[a[i]] = 0;
                    q.push(pii(0, a[i]));
                    cnt++;
                }
            }
            if(cnt == 0)
                break;
            dijkstra();
            for(int i = 0; i < k; i++) {
                if(a[i] & ind)
                    continue;
                if(Min > dis[a[i]])
                    Min = dis[a[i]];
            }
            
            fill(dis, dis + n + 1, INF);
            for(int i = 0; i < k; i++) {
                if(!(a[i] & ind)) {
                    dis[a[i]] = 0;
                    q.push(pii(0, a[i]));
                    cnt++;
                }
            }
            dijkstra();
            for(int i = 0; i < k; i++) {
                if(!(a[i] & ind))
                    continue;
                if(Min > dis[a[i]])
                    Min = dis[a[i]];
            }
            
            ind <<= 1;
        }
        printf("Case #%d: %lld\n", Case++, Min);
    }
    return 0;
}