Choose the best route

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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.

InputThere 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.
OutputThe 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

主要思路：多个起点求最短路径，可以设置一个源点，使得该点到所有起点的长度为0，题目即转化为从一个起点到一个终点的最短路径。（注意，该题用cin读入数据可能会超时，建议用scanf）

#include<iostream>
#include<cstring>
#define INF 99999999
using namespace std;
int dis[1005],visit[1005],e[1005][1005];
int n,m,s;
void dijkstra()
{
    int i,j;
    memset(visit,0,sizeof(visit));
    for (i=1;i<=n;i++)
        dis[i]=e[0][i];
    visit[0]=1;dis[0]=0;
    for (i=0;i<=n;i++)
    {
        int pos=-1,mindis=INF;
        for (j=0;j<=n;j++)
            if (visit[j]==0&&mindis>dis[j])
            {
                pos=j;
                mindis=dis[j];
            }
        if (pos==-1)
            break;
        visit[pos]=1;
        for (j=1;j<=n;j++)
            if (visit[j]==0&&dis[j]>dis[pos]+e[pos][j])
                dis[j]=dis[pos]+e[pos][j];
    }
}
int main()
{
    int p,q,t,i,j,w;
    while (scanf("%d%d%d",&n,&m,&s)!=EOF)
    {
        for (i=0;i<1005;i++)
        {
            for (j=0;j<1005;j++)
                    e[i][j]=INF;
            e[i][i]=0;
        }
        while (m--)
        {
            scanf("%d%d%d",&p,&q,&t);
            if (t<e[p][q])
                e[p][q]=t;
        }
        scanf("%d",&w);
        while (w--)
        {
            scanf("%d",&t);
            e[0][t]=0;
        }
        dijkstra();
        if (dis[s]>=INF)
            cout<<"-1\n";
        else
            cout<<dis[s]<<endl;
    }
    return 0;
}