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;  

}