Network(POJ-1861)

POJ-1861転送ゲート
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
テーマの大意
図、n個の頂点、m個のエッジを指定して、図を連通させ、エッジ重み値と最小値を最大エッジ重みの最小値とエッジの総数を出力し、得られた新しい図を印刷します.
問題を解く構想.
Kruskal:辺権を小さくから大きく並べて、左右の2つの点が1つの集合(すでに連通している)の中にあるかどうかを判断します.
ACコード

#include
#include
using namespace std;
const int maxn = 15100;
const int inf = -999;
int n,m;
int fa[maxn];
int Cnt = 0;
int maxVal = inf;
struct node
{
	int from;
	int to;
	int val;
}Edge[maxn],newEdge[maxn];
bool cmp(node a, node b)
{
	return a.val < b.val;
}
void init()
{
	for(int i = 1 ; i < maxn ; i++)
	{
		fa[i] = i;
	}
}
int Find(int x)
{
	return fa[x] == x? x: fa[x] = Find(fa[x]);
}
void unite(int x ,int y)
{
	fa[Find(x)] = Find(y); 
}
void Kruskal()
{
	for(int i = 1 ; i <= m ; i++)
	{
		int a = Find(Edge[i].from);
		int b = Find(Edge[i].to);
		if(a != b)
		{
			unite(a,b);
			//       ,Cnt      
			Cnt++;
			newEdge[Cnt].from = Edge[i].from;
			newEdge[Cnt].to = Edge[i].to;
			newEdge[Cnt].val = Edge[i].val;	
			//      	
			if(maxVal < Edge[i].val)
			{	
				maxVal = Edge[i].val;
			}
		}
	}
}
void PrintAns()
{
	cout<<maxVal<<endl;
	cout<<Cnt<<endl;
	for(int i = 1 ; i <= Cnt ; i++)
	{
		cout<<newEdge[i].from<<" "<<newEdge[i].to<<endl;
	}
}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>m;
	init();
	for(int i = 1 ; i <= m ; i++)
	{
		cin>>Edge[i].from>>Edge[i].to>>Edge[i].val;
	}
	sort(Edge+1,Edge+m+1,cmp);
	Kruskal();
	PrintAns();
	return 0;
}