2019.2.15 t2

考虑倒过来计算最短路径长度，设dis[u]表示在最坏情况下，点u到最近的一 个出口的最短路，则p个出口的dis值都是0，答案即为dis[0]。

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <queue>
 #include <algorithm>
 using namespace std;

 #define res register int
 #define LL long long
 #define inf 0x3f3f3f3f
 inline int read()
 {
     int x(),f(); char ch;
     while(!isdigit(ch=getchar())) if(ch=='-') f=-;
     while(isdigit(ch)) x=x*+ch-'',ch=getchar();
     return f*x;
 }

 inline int max(int x,int y){return x>y?x:y;}
 inline int min(int x,int y){return x<y?x:y;}
 const int N=+;
 int n,m,k,d,tot;
 int head[N],ver[N<<],nxt[N<<],edge[N<<];
 int dis[N],vis[N];
 inline void add(int x,int y,int z){
     ver[++tot]=y; nxt[tot]=head[x]; head[x]=tot; edge[tot]=z;
 }
 struct node{
     int id;
     LL dd;
     bool operator<(const node &n2) const {
     return dd>n2.dd;}
 };
 priority_queue<node> q;

 inline LL dij()
 {
     while(q.size())
     {
         node now=q.top(); q.pop();
         int x=now.id,z=now.dd;
         if(++vis[x]>d+) continue;
         if(vis[x]==d+ || !dis[x])
         {
             dis[x]=z;
             for(res i(head[x]) ; i ; i=nxt[i])
             {
                 int y=ver[i];
                 if(dis[y]==inf)
                     q.push((node){ver[i],dis[x]+(LL)edge[i]});
             }
         }
     }
 }

 int main()
 {
     freopen("maze.in","r",stdin);
     freopen("maze.out","w",stdout);

     memset(dis,inf,sizeof(dis));
     n=read(); m=read(); k=read(); d=read();
     for(res i= ; i<=m ; i++)
     {
         int x=read(),y=read(),z=read();
         add(x,y,z); add(y,x,z);
     }
     for(res i= ; i<=k ; i++)
     {
         int x=read(); dis[x]=;
         q.push((node){x,});
     }
     dij();
     if(dis[]==inf) puts("-1");
     else cout<<dis[]<<endl;

     return ;
 }