Description
Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input
The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.
Output
Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.
Sample Input
4 6 1 2 1 1 3 1 1 4 2 2 3 1 3 4 1 2 4 1
Sample Output
1 4 1 2 1 3 2 3 3 4
题意：
给你一个N个点和M条边的图,现在要你从这M条边中选一些边的集合,使得单边的长度的最大值最小且所有N个点要连通.要你输出:单边长度的最大值,选的边数目,每条边的两个端点号.
分析：
和找最小生成树一样找到最小生成树输出最大的一条边。
代码：
这个代码，本人在poj上用c++提交错误，换成g++ A了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int maxn=50000;
struct edge
{
int u,v,dis;
edge(int u1,int v1,int dist):u(u1),v(v1),dis(dist){}
edge(){}
};
bool com(edge edge1,edge edge2)
{
return edge1.dis<edge2.dis;
}
struct kruskal
{
int n,m;
edge edges[maxn];
vector<int> e;
int fa[maxn];
int intn(int n)
{
this->n=n;
e.clear();
m=0;
memset(fa,-1,sizeof(fa));
}
int findest(int u)
{
if(fa[u]==-1)return u;
else
{
fa[u]=findest(fa[u]);
}
return fa[u];
}
void addedge(int u,int v,int dis)
{
edges[m++]=edge(u,v,dis);
}
int Kruskal()
{
sort(edges,edges+m,com);
int cnt=0;
for(int i=0;i<m;i++)
{
int u=findest(edges[i].u);
int v=findest(edges[i].v);
if(u!=v)
{
e.push_back(i);
fa[u]=v;
if(++cnt>=n-1)return i;
}
}
return -1;
}
}kk;
int main()
{
int n,m;
int u,v,dis;
cin>>n>>m;
kk.intn(n);
while(m--)
{
cin>>u>>v>>dis;
kk.addedge(u,v,dis);
}
int num=kk. Kruskal();
cout<<kk.edges[num].dis<<endl;
cout<<n-1<<endl;
for(int i=0;i<n-1;i++)
{
int id=kk.e[i];
cout<<kk.edges[id].u<<" "<<kk.edges[id].v<<endl;
}
return 0;
}