#include <stdio.h>

 #include <string.h>

 int map[][];

 int n;

 int Dijkstra(int s,int e){

     bool done[];

     int d[];

     memset(done,,sizeof(done));

     for(int i = ;i <= n;i++)

         d[i] = (i == s ?  : );

     for(int i = ;i <= n;i++){//最多执行n+1次操作

         int minx,minn = ;

         for(int j = ;j <= n;j++)//先找到d[]最小的点

             if(!done[j] && d[j] < minn){

                 minn = d[j];

                 minx = j;

             }

         done[minx] = ;//将该点加入集合

         if(minx == e)

             return d[minx];

         for(int j = ;j <= n;j++){//再更新所有的d[]

             if(!done[j] && d[minx] + map[minx][j] < d[j]){

                 d[j] = d[minx] + map[minx][j];

             }

         }

     }

     return -;//如果没有找到到终点的路径，返回-1

 }

 int main(){

     int m,s;

     int i,j,k;

     int p,q,t;

     int w,ww,ans;

     while(scanf("%d%d%d",&n,&m,&s) != EOF){

         ans = ;

         for(i = ;i < ;i++){

             for(j = ;j < ;j++){

                 if(i == j)

                     map[i][j] = ;

                 else

                     map[i][j] = ;

             }

         }

         while(m--){

             scanf("%d%d%d",&p,&q,&t);

             if(t < map[p][q])//可能两站间存在多条线路取短的那条路

                 map[p][q] = t;

         }

         scanf("%d",&w);

         while(w--){

             scanf("%d",&ww);

             map[][ww] = ;

         }

         ans = Dijkstra(,s);//巧妙之处，加入超级源点0

         if(ans == -)

             printf("-1\n");

         else

             printf("%d\n",ans);

     }

     return ;

 }

One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.

There are several test cases.
Each case begins with three integers
n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number
of bus stations in this city and m stands for the number of directed
ways between bus stations .(Maybe there are several ways between two bus
stations .) s stands for the bus station that near Kiki’s friend’s
home.
Then follow m lines ,each line contains three integers p , q , t
(0<t<=1000). means from station p to station q there is a way and
it will costs t minutes .
Then a line with an integer
w(0<w<n), means the number of stations Kiki can take at the
beginning. Then follows w integers stands for these stations.

The output contains one line for each data set : the least time Kiki
needs to spend ,if it’s impossible to find such a route ,just output
“-1”.

5 8 5

1 2 2

1 5 3

1 3 4

2 4 7

2 5 6

2 3 5

3 5 1

4 5 1

2

2 3

4 3 4

1 2 3

1 3 4

2 3 2

1

1

1

-1

dandelion