题解：由于是多个起点和单个终点，所以反向构图，那么就是多个终点和单个起点了，于是直接最短路。
#include <cstdio>
#include <cstring>
#include <utility>
#include <queue>
using namespace std;
const int N=20005;
const int INF=9999999;
typedef pair<int,int>seg;
priority_queue<seg,vector<seg>,greater<seg> >q;
int begin,end,d[N],head[N],u[N],v[N],w[N],next[N],n,m,a,b,c,k;
bool vis[N];
void build(){
memset(head,-1,sizeof(head));
for(int e=1;e<=m;e++){
scanf("%d%d%d",&v[e],&u[e],&w[e]);
next[e]=head[u[e]]; head[u[e]]=e;
}
}
void Dijkstra(int src){
memset(vis,0,sizeof(vis));
for(int i=0;i<=n;i++) d[i]=INF;
d[src]=0;
q.push(make_pair(d[src],src));
while(!q.empty()){
seg now=q.top(); q.pop();
int x=now.second;
if(vis[x]) continue; vis[x]=true;
for(int e=head[x];e!=-1;e=next[e])
if(d[v[e]]>d[x]+w[e]){
d[v[e]]=d[x]+w[e];
q.push(make_pair(d[v[e]],v[e]));
}
}
}
int main(){
while(~scanf("%d%d%d",&n,&m,&begin)){
build(); int min=INF;
scanf("%d",&k);
Dijkstra(begin);
for(int i=0;i<k;i++){
scanf("%d",&end);
min=d[end]<min?d[end]:min;
}
if(min==INF)puts("-1");
else printf("%d
",min);
}
return 0;
}