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。
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#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,greater >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; } |
HDU 2680 Choose the best route,布布扣,bubuko.com
HDU 2680 Choose the best route
原文：http://www.cnblogs.com/forever97/p/3619101.html