分层最短路（牛客第四场）-- free

题意：

给你边权，起点和终点，有k次机会把某条路变为0，问你最短路是多长。

思路：

分层最短路模板题。题目有点坑（卡掉了SPFA，只能用dijkstra跑的算法）。

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

 const int MAXN=1e3,MAXM=5e3,MAXK=,INF=~0U>>;
 int N,M,K;
 int S,T;

 struct E{int to,next,value;} e[*MAXM*(MAXK+)+MAXK+];int ecnt,G[MAXN*(MAXK+)+];
 void addEdge(int u,int v,int w){e[++ecnt]={v,G[u],w};G[u]=ecnt;}

 void readInt(int & x)
 {
     x=;
     bool flag=false;
     char c;
     do c=getchar(); while(c!='-'&&(c<''||c>''));
     if(c=='-') flag=true;
     else x+=c-'';
     c=getchar();
     while(c>=''&&c<='')
     {
         x*=;
         x+=c-'';
         c=getchar();
     }
 }

 inline void read()
 {
     int i,j;
     readInt(N);readInt(M);
     readInt(S);readInt(T);readInt(K);
     for(i=;i<=M;i++)
     {
         int u,v,w;
         readInt(u);readInt(v);readInt(w);
         for(j=;j<=K;j++)
         {
             addEdge(u+N*j,v+N*j,w);
             addEdge(v+N*j,u+N*j,w);
             if(j!=K)
             {
                 addEdge(u+N*j,v+N*(j+),);
                 addEdge(v+N*j,u+N*(j+),);
             }
         }
     }
 }

 struct HN{int id,v;bool operator<(const HN & ot)const{return v>ot.v;}};

 priority_queue<HN> heap;

 bool inS[MAXN*(MAXK+)+];
 int dis[MAXN*(MAXK+)+];

 void dijkstra(int v0)
 {
     int i;
     for(i=;i<=N*(K+);i++) dis[i]=INF;
     memset(inS,false,sizeof(inS));
     dis[v0]=;
     heap.push((HN){v0,});
     while(!heap.empty())
     {
         int u=heap.top().id;heap.pop();
         if(inS[u]) continue;
         inS[u]=true;
         for(i=G[u];i;i=e[i].next)
         {
             int v=e[i].to;
             if(!inS[v])
                 if(dis[v]>dis[u]+e[i].value)
                 {
                     dis[v]=dis[u]+e[i].value;
                     heap.push((HN){v,dis[v]});
                 }
         }
     }
 }

 int main()
 {
     //freopen("Motor.in","r",stdin);
     //freopen("Motor.out","w",stdout);
     int i;
     read();
     dijkstra(S);
     int ans=INF;
     for(i=;i<=K;i++) ans=min(ans,dis[T+N*i]);
     printf("%d",ans);
     return ;
 }