HDU 2680 Choose the best route

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\n",min);
    }  
    return 0;  
}