[Luogu2901][USACO08MAR]牛慢跑Cow Jogging Astar K短路

题目链接：https://daniu.luogu.org/problem/show?pid=2901

Astar的方程$f(n)=g(n)+h(n)$，在这道题中我们可以反向最短路处理出$h(n)$的精确值。然后跑Astar找K次最短路就好了。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 using namespace std;
 int inline readint(){
     int Num;char ch;
     while((ch=getchar())<''||ch>'');Num=ch-'';
     while((ch=getchar())>=''&&ch<='') Num=Num*+ch-'';
     return Num;
 }
 int N,M,K;
 int to[],ne[],w[],fir[],cnt=;
 void add(int a,int b,int c){
     to[++cnt]=b;
     w[cnt]=c;
     ne[cnt]=fir[a];
     fir[a]=cnt;
 }
 int nto[],nne[],nw[],nfir[],ncnt=;
 void nadd(int a,int b,int c){
     nto[++ncnt]=b;
     nw[ncnt]=c;
     nne[ncnt]=nfir[a];
     nfir[a]=ncnt;
 }
 int dis[];
 bool in[];
 queue <int> q;
 void Spfa(){
     memset(dis,/,sizeof(dis));
     in[]=true;
     dis[]=;
     q.push();
     int u;
     while(!q.empty()){
         int u=q.front();
         q.pop();
         in[u]=false;
         for(int i=nfir[u];i!=-;i=nne[i]){
             int v=nto[i];
             if(dis[v]>dis[u]+nw[i]){
                 dis[v]=dis[u]+nw[i];
                 if(!in[v]){
                     in[v]=true;
                     q.push(v);
                 }
             }
         }
     }
 }
 struct NODE{
     int d,num;
     NODE(int _d=,int _num=){
         d=_d;
         num=_num;
     }
     bool operator < (const NODE &_)const{
         return d>_.d;
     }
 };
 int ans[],rk=;
 priority_queue <NODE> Q;
 void Astar(){
     Q.push(NODE(dis[N],N));
     NODE u;
     while(!Q.empty()){
         u=Q.top();
         Q.pop();
         if(u.num==){
             ans[++rk]=u.d;
             if(rk==K) return;
         }
         for(int i=fir[u.num];i!=-;i=ne[i])
             Q.push(NODE(u.d-dis[u.num]+w[i]+dis[to[i]],to[i]));
     }
 }
 int main(){
     memset(fir,-,sizeof(fir));
     memset(nfir,-,sizeof(nfir));
     N=readint();
     M=readint();
     K=readint();
     for(int i=;i<=M;i++){
         int a=readint(),
             b=readint(),
             c=readint();
         add(a,b,c);
         nadd(b,a,c);
     }
     Spfa();
     Astar();
     for(int i=;i<=K;i++)
         if(ans[i]) printf("%d\n",ans[i]);
         else puts("-1");
     return ;
 }