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
题意：就拿第二个例子来说吧！
                        4（节点数1-4）   3（3条路）   4（源点位置）（起点）
                               1    2    3
                               1    3    4
                               2    3    2；（3条路）
                               1；（一个终点） 
                               1；（终点位置）

    求起点到终点的最短路！！
思路：可以用floyd 直接全部遍历，求出即可！！，但太耗时，可能超时，我这里用dijkstra写的：
 代码：
 
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAX 0x3f3f3f3f
using namespace std;
int map[1010][1010],d[1010],n,m,s,t;
void dijkstra(int x)
{
    int i,j,min,mark,used[1010];
    for(i=1;i<=n;i++)
    {
        used[i]=0;//全部标记为未访问 
        d[i]=map[x][i];//全部指向 源点 
    }
    d[x]=0; //源点到自己为零 
    used[x]=1;//源点标记为访问过 
    for(i=2;i<=n;i++)
    {
        min=MAX;
        mark=-1;
        for(j=1;j<=n;j++)//寻找距离最近的点 
        {
            if(!used[j]&&d[j]<min)
            {
                min=d[j];
                mark=j;//记录下标 
            }
        }
        if(mark==-1)
        break;
        used[mark]=1;//将下标的节点标记为已访问 
        for(j=1;j<=n;j++)
        {
            if(!used[j]&&d[j]>d[mark]+map[mark][j])//更新路径 
            d[j]=d[mark]+map[mark][j];
        }
    }

}
int main()
{
    int a,b,c,i,j;
    while(scanf("%d%d%d",&n,&m,&s)!=EOF)
    {
        memset(map,MAX,sizeof(map));
        for(i=0;i<m;i++)//取图 
        {
            scanf("%d%d%d",&a,&b,&c);
            if(map[b][a]>c)//防止重边 
               map[b][a]=c;
        }
        dijkstra(s);
        int start;
        int mi=MAX;
        scanf("%d",&t);
        for(i=0;i<t;i++)
        {
            scanf("%d",&start);
            mi=mi<d[start]?mi:d[start];//求出最短距离 
        }
        if(mi==MAX)//无法找到最短路 
        printf("-1\n");
        else
        printf("%d\n",mi);
    }
return 0;
}