Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of the cables required to connect his farm to the phone system.

There are N (1 ≤ N ≤ 1,000) forlorn telephone poles conveniently numbered 1..N that are scattered around Farmer John‘s property; no cables connect any them. A total of P (1 ≤ P ≤ 10,000) pairs of poles can be connected by a cable; the rest are too far apart.

The i-th cable can connect the two distinct poles A_i and B_i, with length L_i (1 ≤ L_i ≤ 1,000,000) units if used. The input data set never names any {A_i, B_i} pair more than once. Pole 1 is already connected to the phone system, and pole N is at the farm. Poles 1 and N need to be connected by a path of cables; the rest of the poles might be used or might not be used.

As it turns out, the phone company is willing to provide Farmer John with K (0 ≤ K < N) lengths of cable for free. Beyond that he will have to pay a price equal to the length of the longest remaining cable he requires (each pair of poles is connected with a separate cable), or 0 if he does not need any additional cables.

Determine the minimum amount that Farmer John must pay.

  1 #include<iostream>  2 #include<cstdio>  3 #include<cstdlib>  4 #include<cctype>  5 #include<cmath>  6 #include<queue>  7 #include<cstring>  8 #include<map>  9 #include<stack> 10 #include<set> 11 #include<vector> 12 #include<algorithm> 13 #include<string.h> 14 typedef long long ll; 15 typedef unsigned long long LL; 16 using namespace std; 17 const int INF=0x3f3f3f3f; 18 const double eps=0.0000000001; 19 const int N=100000+10; 20 struct node{ 21     int to,next,w; 22     bool operator<(const node &a)const{ 23         return w>a.w; 24     } 25 }edge[N*2]; 26 int t; 27 int head[N]; 28 int vis[N],dis[N]; 29 int n; 30 void init(){ 31     memset(vis,0,sizeof(vis)); 32     memset(head,-1,sizeof(head)); 33     t=0; 34     for(int i=0;i<N;i++)dis[i]=INF; 35 } 36 void add(int u,int v,int w){ 37     edge[t].to=v; 38     edge[t].w=w; 39     edge[t].next=head[u]; 40     head[u]=t++; 41 } 42 int Dijkstra(int x,int y){ 43     memset(dis,INF,sizeof(dis)); 44     dis[x]=0; 45     node t1,t2; 46     t1.to=x; 47     priority_queue<node>q; 48     q.push(t1); 49     memset(vis,0,sizeof(vis)); 50     while(!q.empty()){ 51         //cout<<2<<endl; 52         t1=q.top(); 53         q.pop(); 54         int u=t1.to; 55         if(vis[u]==1)continue; 56         vis[u]=1; 57         for(int i=head[u];i!=-1;i=edge[i].next){ 58             int v=edge[i].to; 59             int tt=0; 60             if(edge[i].w>=y)tt=1; 61             if(vis[v]==0&&dis[v]>dis[u]+tt){ 62                 dis[v]=dis[u]+tt; 63                 t2.to=v; 64                 t2.w=dis[v]; 65                 q.push(t2); 66             } 67         } 68     } 69     return dis[n]; 70 } 71 int main(){ 72     int k,p; 73     while(scanf("%d%d%d",&n,&p,&k)!=EOF){ 74         init(); 75         int maxx=0; 76         for(int i=0;i<p;i++){ 77             int u,v,w; 78             scanf("%d%d%d",&u,&v,&w); 79             add(u,v,w); 80             add(v,u,w); 81             maxx=max(maxx,w); 82         } 83         int low=0; 84         int high=maxx; 85         int ans=0; 86         while(low<=high){ 87             int mid=(low+high)>>1; 88             //cout<<Dijkstra(1,mid)<<endl; 89             if(Dijkstra(1,mid)<=k){ 90                 high=mid-1; 91             } 92             else{ 93                 ans=mid; 94                 low=mid+1; 95             } 96  97         } 98         if(low>maxx){cout<<-1<<endl;continue;} 99         cout<<ans<<endl;100         //for(int i=1;i<=n;i++)cout<<dis[i]<<endl;101     }102 }