Problem Description 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.  

Input 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.

Output 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”.  

Sample Input

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  

Sample Output

1 -1  

题意：给出n,m,s，代表有n个车站，m条路线，s为终点

然后m行代表路线和花费，注意是单向边

最后给出k个数代表能选择的出发点

思路：这道题其实和HDU2066一个人的旅行的题目大同小异，将0看成出发点，将n个能选择的点看做事0的相邻并且花费为0的点，那么就是求0到S的最小花费了

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int inf = 100000000;
int n,m,s;
int map[1005][1005];
int vis[1005],cast[1005];

void Dijkstra()
{
    int i,j,minn,pos;
    memset(vis,0,sizeof(vis));
    vis[0] = 1;
    for(i = 0; i<=n; i++)
        cast[i] = map[0][i];
    for(i = 0; i<=n; i++)
    {
        minn = inf;
        for(j = 0; j<=n; j++)
        {
            if(cast[j]<minn && !vis[j])
            {
                pos = j;
                minn = cast[j];
            }
        }
        vis[pos] = 1;
        for(j = 0; j<=n; j++)
        {
            if(cast[pos]+map[pos][j]<cast[j] && !vis[j])
                cast[j] = cast[pos]+map[pos][j];
        }
    }
}


int main()
{
    int i,j;
    int x,y,t;
    while(~scanf("%d%d%d",&n,&m,&s))
    {
        for(i = 0; i<=n; i++)
            for(j = 0; j<=n; j++)
                map[i][j] = inf;
        while(m--)
        {
            scanf("%d%d%d",&x,&y,&t);
            if(t<map[x][y])
            map[x][y] = t;
        }
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d",&x);
            map[0][x] = 0;
        }
        Dijkstra();
        if(cast[s] == inf)
            printf("-1\n");
        else
            printf("%d\n",cast[s]);
    }

    return 0;
}