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 
     
      #include 
      
       #include 
       
        using namespace std;#define MAX 1005#define INF 1<<30int 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
        
         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 <