P5304 [GXOI/GZOI2019]旅行者

链接；https://www.luogu.com.cn/problem/P5304

题目描述：给定一个\(n\)个点\(m\)条边的有向图，求\(k\)个点中任意两点的最短路。

题解:由于有\(k\)个关键点，我们可以把它看作\(「AMPPZ2014」Petrol\)的"加油站"，于是我们可以跑正反两图最短路然后算将每一个点的贡献即可。

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
struct node
{
  int v,data,nxt;
};
node edge[2][1000001];
long long T,n,m,K,x,y,res,in[100001],pv[2][100001],head[2][100001],d[100001],len[2],dis[2][100001];
bool used[100001];
void add(int k,int x,int y,int z)
{
  edge[k][++len[k]].v=y;
  edge[k][len[k]].data=z;
  edge[k][len[k]].nxt=head[k][x];
  head[k][x]=len[k];
  return;
}
int read()
{
  char c=0;
  int sum=0;
  while (c<'0'||c>'9')
    c=getchar();
  while ('0'<=c&&c<='9')
    {
      sum=sum*10+c-'0';
      c=getchar();
    }
  return sum;
}
struct reads
{
  long long num,data;
  bool operator < (const reads &a)const
  {
    return data>a.data;
  }
};
reads tmp;
reads make_reads(int x,int y)
{
  tmp.num=x;
  tmp.data=y;
  return tmp;
}
void dijkstra(int k)
{
  priority_queue<reads>q;
  int top;
  for (int i=1;i<=n;++i)
    dis[k][i]=1e18,used[i]=0;
  for (int i=1;i<=K;++i)
    {
      dis[k][d[i]]=0;
      pv[k][d[i]]=d[i];
      q.push(make_reads(d[i],0));
    }
  while (!q.empty())
    {
      top=q.top().num;
      q.pop();
      if (used[top])
	continue;
      used[top]=1;
      for (int i=head[k][top];i>0;i=edge[k][i].nxt)
	if (dis[k][edge[k][i].v]>dis[k][top]+edge[k][i].data)
          {
	    dis[k][edge[k][i].v]=dis[k][top]+edge[k][i].data;
	    pv[k][edge[k][i].v]=pv[k][top];
	    q.push(make_reads(edge[k][i].v,dis[k][edge[k][i].v]));
	  }
    }
  return;
}
int main()
{
  int x,y,z;
  T=read();
  while (T--)
    {
      res=len[0]=len[1]=0;
      for (int i=1;i<=n;++i)
	head[0][i]=head[1][i]=0;
      n=read(),m=read(),K=read();
      for (int i=1;i<=m;++i)
	{
	  x=read(),y=read(),z=read();
	  add(0,x,y,z);
	  add(1,y,x,z);
	}
      for (int i=1;i<=K;++i)
	d[i]=read();
      dijkstra(0),dijkstra(1);
      res=1e18;
      for (int i=1;i<=n;++i)
	for (int j=head[0][i];j>0;j=edge[0][j].nxt)
	  if (pv[0][i]!=pv[1][edge[0][j].v])
	      res=min(res,dis[0][i]+dis[1][edge[0][j].v]+edge[0][j].data);
      printf("%lld\n",res);
    }
  return 0;
}