本以为是个二进制分组傻逼题https://www.cnblogs.com/Gloid/p/9545753.html，实际上有神仙的一个log做法https://www.cnblogs.com/asuldb/p/10721251.html。下面代码是二进制分组的。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
#define N 100010
#define M 500010
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,k,p[N],a[N],t;
ll d[N];
bool flag[N];
struct data{int to,nxt,len;
}edge[M];
void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
struct data2
{
int x;ll d;
bool operator <(const data2&a) const
{
return d>a.d;
}
};
priority_queue<data2> q;
void dijkstra()
{
memset(flag,0,sizeof(flag));
for (;;)
{
while (!q.empty()&&flag[q.top().x]) q.pop();
if (q.empty()) break;
data2 x=q.top();q.pop();
flag[x.x]=1;
for (int i=p[x.x];i;i=edge[i].nxt)
if (x.d+edge[i].len<d[edge[i].to])
{
d[edge[i].to]=x.d+edge[i].len;
q.push((data2){edge[i].to,d[edge[i].to]});
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj5506.in","r",stdin);
freopen("bzoj5506.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
int T=read();
while (T--)
{
n=read(),m=read(),k=read();
memset(p,0,sizeof(p));t=0;
for (int i=1;i<=m;i++)
{
int x=read(),y=read(),z=read();
addedge(x,y,z);
}
for (int i=1;i<=k;i++) a[i]=read();
ll ans=10000000000000000ll;
for (int _=18;~_;_--)
{
while (!q.empty()) q.pop();
memset(d,42,sizeof(d));
for (int i=1;i<=k;i++)
if (a[i]&(1<<_)) d[a[i]]=0,q.push((data2){a[i],0});
dijkstra();
for (int i=1;i<=k;i++)
if (!(a[i]&(1<<_))) ans=min(ans,d[a[i]]);
while (!q.empty()) q.pop();
memset(d,42,sizeof(d));
for (int i=1;i<=k;i++)
if (!(a[i]&(1<<_))) d[a[i]]=0,q.push((data2){a[i],0});
dijkstra();
for (int i=1;i<=k;i++)
if (a[i]&(1<<_)) ans=min(ans,d[a[i]]);
}
cout<<ans<<endl;
}
return 0;
}
BZOJ5506 GXOI/GZOI2019旅行者（最短路）
原文：https://www.cnblogs.com/Gloid/p/10725368.html