Poj3662 telephone lines binary + Shortest Path

 

Telephone Lines

Time Limit:1000 ms		Memory Limit:65536 K
Total Submissions:3468		Accepted:1245

Description

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

TheI-Th cable can connect the two distinct polesAIAndBi, With length
Li(1 ≤Li≤ 1,000,000) units if used. The input data set never names any {AI,Bi} Pair more than once. Pole 1 is already connected to the phone system, and poleNIs
The farm. Poles 1 andNNeed 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
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.

Input

* Line 1: Three space-separated integers:N,P, AndK
* Lines 2 ..P+ 1: LineI+ 1 contains the three space-separated integers:AI,
Bi, AndLi

Output

* Line 1: A single integer, the minimum amount Farmer John can pay. If it is impossible to connect the farm to the phone company, print-1.

Sample Input

5 7 11 2 53 1 42 4 83 2 35 2 93 4 74 5 6

Sample output

4

Source

USACO 2008 January Silver binary + Shortest Path: Calculate a path from 1 to n to minimize the side of k + 1. If the original edge weight is less than or equal to the value of the new Edge Weight of the mid is 0, otherwise the new edge weight is 1. Find the shortest path. If the value is less than or equal to k, the condition is met. Code:

#include<cstdio>#include<cstring>#define N 1005int n,m,k,num,adj[N],low[N],f[N],q[N];struct edge{int v,w,c,pre;}e[N*20];void insert(int u,int v,int w){e[num].v=v;e[num].c=w;e[num].pre=adj[u];adj[u]=num++;}int spfa(int x){int i,v,head=0,tail=0;memset(f,0,sizeof(f));memset(low,0x7f,sizeof(low));low[x]=0;q[++tail]=x;while(head!=tail){x=q[head=(head+1)%N];f[x]=0;for(i=adj[x];~i;i=e[i].pre)if(low[v=e[i].v]>low[x]+e[i].w){low[v]=low[x]+e[i].w;if(!f[v]){f[v]=1;q[tail=(tail+1)%N]=v;}}}return low[n]<=k;}int ok(int x){int i,j;for(i=1;i<=n;i++)for(j=adj[i];~j;j=e[j].pre)if(e[j].c<=x)e[j].w=0;elsee[j].w=1;return spfa(1);}int main(){int u,v,w;while(~scanf("%d%d%d",&n,&m,&k)){num=0;memset(adj,-1,sizeof(adj));int l=0,r=0,mid,ans=-1;while(m--){scanf("%d%d%d",&u,&v,&w);insert(u,v,w);insert(v,u,w);if(w>r)r=w;}while(l<=r){mid=(l+r)/2;if(ok(mid)){ans=mid;r=mid-1;}elsel=mid+1;}printf("%d\n",ans);}}