题意：给你k个点，让你求两两最短路之间的最小值
我们考虑二进制拆分，使得每两个点都有机会分在不同的组((A:0,B:1))中，从源点(S)向(A/B)中的点连边权为0的边，从(B/A)中的点向汇点(T)连边权为0的边，这时(S->T)的最短路就是(A/B)中的点到(B/A)中的点最短路的最小值
所以做最短路次数为(2log k)，总复杂度为(T n log nlog k)（srf好像还有少一个log的做法，orz srf）
#include <bits/stdc++.h>
#define ll long long
#define N 100005
#define M 700005
#define getchar nc
using namespace std;
inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int read()
{
register int x=0,f=1;register char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return x*f;
}
inline void write(register ll x)
{
if(!x)putchar('0');if(x<0)x=-x,putchar('-');
static int sta[20];register int tot=0;
while(x)sta[tot++]=x%10,x/=10;
while(tot)putchar(sta[--tot]+48);
}
inline ll Min(register ll a,register ll b)
{
return a<b?a:b;
}
struct edge{
int to,next,w;
}e[M];
int head[N],cnt,headn[N],cntn;
inline void add(register int u,register int v,register int w)
{
e[++cnt]=(edge){v,head[u],w};
head[u]=cnt;
}
inline void addn(register int u,register int v,register int w)
{
e[++cntn]=(edge){v,headn[u],w};
headn[u]=cntn;
}
int T,n,m,k,p[N],s,t;
ll dis[N];
inline ll dijkstra()
{
memset(dis,0x3f,sizeof(dis));
dis[s]=0;
priority_queue<pair<ll,int> >q;
q.push(make_pair(0,s));
for(register int i=1;i<n+2;++i)
{
while(!q.empty()&&dis[q.top().second]!=-q.top().first)
q.pop();
if(q.empty())
break;
int now=q.top().second;
q.pop();
for(register int i=headn[now];i;i=e[i].next)
{
int v=e[i].to;
if(dis[v]>dis[now]+e[i].w)
q.push(make_pair(-(dis[v]=dis[now]+e[i].w),v));
}
}
return dis[t];
}
int main()
{
int T=read();
while(T--)
{
cnt=0;
memset(head,0,sizeof(head));
n=read(),m=read(),k=read();
for(register int i=1;i<=m;++i)
{
int u=read(),v=read(),w=read();
add(u,v,w);
}
for(register int i=1;i<=k;++i)
p[i]=read();
ll ans=~0ull>>1;
s=n+1,t=n+2;
for(register int i=0;(1<<i)<=k;++i)
{
cntn=cnt;
memcpy(headn,head,sizeof(head[0])*(n+3));
for(register int j=1;j<=k;++j)
if(j&(1<<i))
addn(p[j],t,0);
else
addn(s,p[j],0);
ans=Min(ans,dijkstra());
cntn=cnt;
memcpy(headn,head,sizeof(head[0])*(n+3));
for(register int j=1;j<=k;++j)
if(j&(1<<i))
addn(s,p[j],0);
else
addn(p[j],t,0);
ans=Min(ans,dijkstra());
}
write(ans),puts("");
}
return 0;
}