枚举全部相邻城市，作为起点，多次spfa，然后每次在想去的城市中找出spfa后的距离起点最短的花费时间

#include <iostream>

#include <cstring>

#include <queue>

using namespace std;

#define MAX 1005

#define INF 1<<30

int T,S,D;

struct Edge{

    int to,time,next;

}edge[MAX*2];

int head[MAX],tol;

int s_city[MAX],d_city[MAX];

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

{

    edge[tol].to = v;

    edge[tol].time = time;

    edge[tol].next = head[u];

    head[u] = tol ++;

}

int dis[MAX];

bool flag[MAX];

int spfa(int src)

{

    for(int i = 0; i < MAX; i ++) dis[i] = INF;

    memset(flag,false,sizeof(flag));

    flag[src] = true;

    dis[src] = 0;

    queue<int>q;

    q.push(src);

    while(!q.empty())

    {

        int u = q.front(); q.pop();

        flag[u] = false;

        for(int i = head[u]; i != -1; i = edge[i].next)

        {

            int v = edge[i].to,time = edge[i].time;

            if(dis[u] + time < dis[v])

            {

                dis[v] = dis[u]+time;

                if(!flag[v]){

                    q.push(v); flag[v] = true;

                }

            }

        }

    }

    int ans = INF;

    for(int i = 0; i < D; i ++)

    {

        if(dis[d_city[i]] < ans) ans = dis[d_city[i]];

    }

    return ans;

}

int main()

{

    int a,b,time;

    while(cin >> T >> S >> D)

    {

        memset(head,-1,sizeof(head));

        tol = 0;

        while(T--){

            cin >> a >> b >> time;

            add(a,b,time); add(b,a,time);

        }

        for(int i = 0; i < S; i ++) cin >> s_city[i];

        for(int i = 0; i < D; i ++) cin >> d_city[i];

        //ans 无穷大。对于每个相邻城市作为源点spfa，而且返回到D个想去的城市中的最小时间

        int ans = INF;

        for(int i = 0; i < S; i ++)

        {

            int temp = spfa(s_city[i]);

            if(temp < ans) ans = temp;

        }

        cout << ans <<endl;

    }

    return 0;

}