Telephone Lines

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<algorithm>  3 #include<queue>  4 #include<stack>  5 #include<cmath>  6 #include<string.h>  7 #include<stdio.h>  8 #include<stdlib.h>  9 #include <set> 10 #include <map> 11 using namespace std; 12 inline int read() 13 { 14     int x=0,f=1; 15     char ch=getchar(); 16     while(ch<‘0‘||ch>‘9‘) 17     { 18         if(ch==‘-‘) 19             f=-1; 20         ch=getchar(); 21     } 22     while(ch>=‘0‘&&ch<=‘9‘) 23     { 24         x=x*10+ch-‘0‘; 25         ch=getchar(); 26     } 27     return x*f; 28 } 29 inline void write(int x) 30 { 31     if(x<0) 32     { 33         putchar(‘-‘); 34         x=-x; 35     } 36     if(x>9) 37         write(x/10); 38     putchar(x%10+‘0‘); 39 } 40 struct Node 41 { 42     int id,dis; 43 }; 44 bool operator<(const Node &a,const Node &b) 45 { 46     return a.dis>b.dis; 47 } 48 typedef struct 49 { 50     int v,next,cost; 51 }Edge; 52 Edge e[50050]; 53 int head[1010]; 54 int d[1010]; 55 int n,p,k; 56 int cnt; 57 inline bool BFS() 58 { 59     int vis[1010]; 60     memset(vis,false,sizeof(vis)); 61     queue<int>Q; 62     Q.push(1); 63     vis[1]=1; 64     while(!Q.empty()) 65     { 66         int u=Q.front(); 67         Q.pop(); 68         for(int i=head[u];i!=-1;i=e[i].next) 69         { 70             int v=e[i].v; 71             if(vis[v]) 72                 continue; 73             vis[v]=1; 74             if(v==n) 75                 return 1; 76             Q.push(v); 77         } 78     } 79     return 0; 80 } 81 inline int Dijkstra(int ans) 82 { 83     priority_queue<Node>q; 84     Node temp,pp; 85     bool vis[1010]; 86     for(int i=1;i<=n;i++) 87     { 88         d[i]=1e+8; 89         vis[i]=0; 90     } 91     d[1]=0; 92     temp.id=1; 93     temp.dis=0; 94     q.push(temp); 95     while(!q.empty()) 96     { 97         pp=q.top(); 98         q.pop(); 99         int u=pp.id;100         if(vis[u])101             continue;102         vis[u]=1;103         for(int i=head[u];i!=-1;i=e[i].next)104         {105             int v=e[i].v;106             int t=e[i].cost>ans?true:false;107             if(!vis[v])108             {109                 if(d[v]>d[u]+t)110                 {111                     d[v]=d[u]+t;112                     pp.id=v;113                     pp.dis=d[v];114                     q.push(pp);115                 }116             }117         }118     }119     return d[n];120 }121 int main()122 {123     int fr,to,cost;124     while(scanf("%d%d%d",&n,&p,&k)!=EOF)125     {126         cnt=1;127         memset(e,0,sizeof(e));128         memset(head,-1,sizeof(head));129         for(int i=0;i<p;i++)130         {131             fr=read();132             to=read();133             cost=read();134             e[cnt].v=to;135             e[cnt].next=head[fr];136             e[cnt].cost=cost;137             head[fr]=cnt;138             cnt++;139             e[cnt].v=fr;140             e[cnt].next=head[to];141             e[cnt].cost=cost;142             head[to]=cnt;143             cnt++;144         }145         if(!BFS())146         {147             puts("-1");148             continue;149         }150         int l=0,r=1000010,mid;151         while(l<=r)152         {153             mid=(l+r)/2;154             if(Dijkstra(mid)<=k)155                 r=mid-1;156             else157                 l=mid+1;158         }159         write(r+1);160         printf("\n");161     }162     return 0;163 }