题面
题解
以所有的感兴趣的城市为起点，我们正着和反着各跑一边多源最短路。记(c_{0/1,i})分别表示正图/反图中离(i)最近的起点，那么对于每条边((u,v,w))，如果(c_{0,u} eq c_{1,v})，那么我们就用(d_{0,u}+d_{1,v}+w)更新答案
//minamoto
#include<bits/stdc++.h>
#define R register
#define ll long long
#define inline __inline__ __attribute__((always_inline))
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=1e5+5,M=5e5+5;
struct eg{int v,nx,w;}e[M];int head[N],tot;
inline void add(R int u,R int v,R int w){e[++tot]={v,head[u],w},head[u]=tot;}
struct EG{int u,v,w;}E[M];int c[2][N],st[N],n,m,k;ll d[2][N],res;bool vis[N];
struct node{
int u;ll d;
inline node(R int uu,R ll dd):u(uu),d(dd){}
inline bool operator <(const node &b)const{return d>b.d;}
};priority_queue<node>q;
void dij(int *c,ll *d){
memset(d,0x3f,(n+1)<<3),memset(c,0,(n+1)<<2),memset(vis,0,n+1);
fp(i,1,k)q.push(node(st[i],d[st[i]]=0)),c[st[i]]=st[i];
while(!q.empty()){
int u=q.top().u;q.pop();if(vis[u])continue;
go(u)if(cmin(d[v],d[u]+e[i].w))q.push(node(v,d[v])),c[v]=c[u];
}
}
int main(){
// freopen("testdata.in","r",stdin);
for(int T=read();T;--T){
n=read(),m=read(),k=read(),res=1e18;
for(R int i=1,u,v,w;i<=m;++i)u=read(),v=read(),w=read(),E[i]={u,v,w},add(u,v,w);
fp(i,1,k)st[i]=read();
dij(c[0],d[0]);
memset(head,0,(n+1)<<2),tot=0;
fp(i,1,m)add(E[i].v,E[i].u,E[i].w);
dij(c[1],d[1]);
memset(head,0,(n+1)<<2),tot=0;
for(R int i=1,u,v;i<=m;++i){
u=E[i].u,v=E[i].v;
if(c[0][u]&&c[1][v]&&c[0][u]!=c[1][v])cmin(res,d[0][u]+d[1][v]+E[i].w);
}
printf("%lld
",res);
}
return 0;
}