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2013
11-10

POJ 1861 Network [解题报告] Java

Network

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1

样例输出:

1
4
1 2
1 3
2 3
3 4

解题代码:

//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;
class Edge implements Comparable{
	int u,v,disten;
	Edge(){};
	void set(int u,int v,int disten){
		this.u = u;
		this.v = v;
		this.disten = disten;
	}
	public int compareTo(Object obj){
		Edge temp = (Edge)obj;
		if(this.disten>temp.disten) return 1;
		return -1;
	}
}
class Set{
	int father[],num[];
	void init(int n){
		num = new int[n+10];
		father = new int[n+10];
		Arrays.fill(num, 1);
		Arrays.fill(father, -1);
	}
	int Ufind(int who){
		int temp,cnt = who;
		while(father[who]!=-1){
			who = father[who];
		}
		while(cnt!=who){
			temp = father[cnt];
			father[cnt] = who;
			cnt = temp;
		}
		return who;
	}
	void Uset(int u,int v){
		if(u==v) return ;
		if(num[u]>num[v]){
			father[v] = u;
			num[u]+=num[v];
		}else{
			father[u] = v;
			num[v]+=num[u];
		}
	}
}
class Kruscal{
	final int N = 15000+100;
	int n,m,cnt,max,ans;
	Edge edge[] = new Edge[N];
	Kruscal(){
		for(int i=0;i< N;++i) edge[i] = new Edge();
	}
	void init(int n,int m){
		cnt = 0;
		this.n = n;
		this.m = m;
		max = ans = 0;
	}
	void addEdge(int u,int v,int disten){
		edge[cnt++].set(u, v, disten);
	}
	void calc(PrintWriter out){
		Set set = new Set();
		set.init(n);
		Arrays.sort(edge,0,m);
		for(int i=0;i< m;++i){
			if(set.Ufind(edge[i].u) == set.Ufind(edge[i].v)){
				edge[i].disten = -1;
			}else{
				max = Max(max,edge[i].disten);
				++ans;
				set.Uset(set.Ufind(edge[i].u), set.Ufind(edge[i].v));
			}
		}
		
		out.println(max);
		out.println(ans);
		for(int i=0;i< m;++i)if(edge[i].disten>=0){
			out.println(edge[i].u+" "+edge[i].v);
		}
	}
	int Max(int a,int b){
		return a>b?a:b;
	}
}
public class Main {
 public static void main(String[]args)throws Exception{
  StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	Kruscal kruscal = new Kruscal();
	int n,m,u,v,disten;
	n = GetNum(cin);
	m = GetNum(cin);
	kruscal.init(n, m);
	for(int i=0;i< m;++i){
		u = GetNum(cin);
		v = GetNum(cin);
		disten = GetNum(cin);
		if(u>v) kruscal.addEdge(v, u, disten);
		else kruscal.addEdge(u, v, disten);
	}
	kruscal.calc(out);
	out.flush();
 }
	static int GetNum(StreamTokenizer cin)throws Exception{
		cin.nextToken();
		return (int) cin.nval;
	}
}

  1. 第一题是不是可以这样想,生了n孩子的家庭等价于n个家庭各生了一个1个孩子,这样最后男女的比例还是1:1

  2. 老实说,这种方法就是穷举,复杂度是2^n,之所以能够AC是应为题目的测试数据有问题,要么数据量很小,要么能够得到k == t,否则即使n = 30,也要很久才能得出结果,本人亲测