逃跑

题解：

 

以下是三次bfs的AC程序

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct Messi
{
    int next,to;
    long long dis1,dis2;
}edge[300001];
int head[100005],num;
long long d1[100005],d2[100005],d3[100005];
int vis[100005];
long long ans,ans0,q[4000005],n,m,k,kk[100005],f[100005];
void add(int u,int v,long long dis1,long long dis2)
{
    num++;
    edge[num].next=head[u];
    head[u]=num;
    edge[num].to=v;
    edge[num].dis1=dis1;
    edge[num].dis2=dis2;
}
void bfs1(int S)
{int h,t;
    memset(vis,0,sizeof(vis));
    memset(d1,127/3,sizeof(d1));
   h=0;t=1;
   q[1]=S;
   d1[S]=0;
    while (h<t)  
    {
        h++;
        h%=4000000;
         int u=q[h];
         vis[u]=0;
          for (int i=head[u];i!=-1;i=edge[i].next)
          {
            int v=edge[i].to;
             if (d1[u]+edge[i].dis1<d1[v])
             {
                d1[v]=d1[u]+edge[i].dis1;
                if (!vis[v])
                {
                 t++;
                 t%=4000000;
                 q[t]=v;
                 vis[v]=1;  
                }
             }
          }
    }
}
void bfs2(int S)
{int h,t,i;
    memset(vis,0,sizeof(vis));
    memset(d2,127/3,sizeof(d2));
   h=0;t=1;
   q[1]=S;
   d2[S]=0;
    while (h<t)  
    {
        h++;
        h%=4000000;
         int u=q[h];
         vis[u]=0;
          for (i=head[u];i!=-1;i=edge[i].next)
          {
            int v=edge[i].to;
             if (d2[u]+edge[i].dis2<d2[v])
             {
                d2[v]=d2[u]+edge[i].dis2;
                if (!vis[v])
                {
                 t++;
                 t%=4000000;
                 q[t]=v;
                 vis[v]=1;  
                }
             }
          }
    }
}
void bfs3(int S)
{int h,t,i;
    memset(vis,0,sizeof(vis));
    memset(d3,127/3,sizeof(d3));
   h=0;t=1;
   q[1]=S;
   d3[S]=0;
    while (h<t)  
    {
        h++;
        h%=4000000;
         int u=q[h];
         vis[u]=0;
          for (i=head[u];i!=-1;i=edge[i].next)
          {
            int v=edge[i].to;
             if (d3[u]+edge[i].dis1<d3[v])
             {
                 d3[v]=d3[u]+edge[i].dis1;
                if (!vis[v])
                {
                 t++;
                 t%=4000000;
                 q[t]=v;
                 vis[v]=1;  
                }
             }
          }
    }
}
int main()
{int S,T,i,u,v;
long long dis1,dis2;
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
    cin>>n>>m>>k>>S>>T;
    memset(head,-1,sizeof(head));
    for (i=1;i<=k;i++)
    {
        scanf("%d",&kk[i]);
    }
     for (i=1;i<=m;i++)
     {
        scanf("%d%d%lld%lld",&u,&v,&dis1,&dis2);
         add(u,v,dis1,dis2);
         add(v,u,dis1,dis2);
     }
    bfs1(T); 
     for (i=1;i<=k;i++)
      ans0+=d1[kk[i]];
    bfs2(S);
     for (i=1;i<=k;i++)
     {
        int x=kk[i];
         add(0,x,-d1[x],0);
     }
     bfs3(0);
     long long ans=2e15;
     for (i=1;i<=n;i++)
      {
       f[i]=d1[i]+d2[i]+d3[i];
      if (f[i]<ans) ans=f[i];
     }
    cout<<ans+ans0;
}