POJ 3662 Telephone Lines

Telephone Lines

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7210		Accepted: 2634

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 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.

Input

* Line 1: Three space-separated integers: N, P, and K
* Lines 2..P+1: Line i+1 contains the three space-separated integers: A_i, B_i, and L_i

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 1
1 2 5
3 1 4
2 4 8
3 2 3
5 2 9
3 4 7
4 5 6

Sample Output

4

最短路+二分

#include <iostream> 
#include <algorithm> 
#include <cstring> 
#include <cstdio>
#include <vector> 
#include <queue> 
#include <cstdlib> 
#include <iomanip>
#include <cmath> 
#include <ctime> 
#include <map> 
#include <set> 
using namespace std; 
#define lowbit(x) (x&(-x)) 
#define max(x,y) (x>y?x:y) 
#define min(x,y) (x<y?x:y) 
#define MAX 100000000000000000 
#define MOD 1000000007
#define pi acos(-1.0) 
#define ei exp(1) 
#define PI 3.141592653589793238462
#define ios() ios::sync_with_stdio(false)
#define INF 0x3f3f3f3f3f 
#define mem(a) (memset(a,0,sizeof(a))) 
typedef long long ll;
struct Node
{
    int x;
    int y;
    int l;
    friend bool operator<(const Node &a,const Node &b)
    {
        return a.l<b.l;
    }
}node[100006];
int a[10006][10006];
int dis[10006];
int vis[10006],n,m,k,x,y,z;
void build(int l)
{
    for(int i=0;i<=n;i++)
    {
        for(int j=0;j<=n;j++)
        {
            a[i][j]=MOD;
        }
        a[i][i]=0;
    }
    for(int i=0;i<m;i++)
    {
        if(node[i].l<=l)
            a[node[i].x][node[i].y]=a[node[i].y][node[i].x]=0;
        else a[node[i].x][node[i].y]=a[node[i].y][node[i].x]=1;
    }
}
int disj(int u)
{
    for(int i=1;i<=n;i++)
    {
        dis[i]=a[i][u];
        vis[i]=0;
    }
    int v,maxn;
    vis[u]=1;
    for(int i=2;i<=n;i++)
    {
        maxn=MOD;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && dis[j]<maxn)
            {
                v=j;
                maxn=dis[j];
            }
        }
        if(maxn==MOD) break;
        vis[v]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j] && dis[j]>dis[v]+a[v][j])
            {
                dis[j]=dis[v]+a[v][j];
            }
        }
    }
    return dis[n];
}
int bin(int k)
{
    int l=0,r=m-1;
    while(l<=r)
    {
        int mid=l+r>>1;
        build(node[mid].l);
        if(disj(1)<=k) r=mid-1;
        else l=mid+1;
    }
    return l;
}
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i=0;i<m;i++)
    {
        scanf("%d%d%d",&node[i].x,&node[i].y,&node[i].l);
    }
    sort(node,node+m);
    build(0);
    int ans=disj(1);
    if(ans==MOD) printf("-1
");
    else
    {
        if(ans<=k) printf("0
");
        else printf("%d
",node[bin(k)].l);
    }
    return 0;
}