首页 > 专题系列 > Java解POJ > POJ 1287 Networking [解题报告] Java
2013
11-09

POJ 1287 Networking [解题报告] Java

Networking

问题描述 :

You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.

Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.

输入:

The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.

The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.

输出:

For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.

样例输入:

1 0

2 3
1 2 37
2 1 17
1 2 68

3 7
1 2 19
2 3 11
3 1 7
1 3 5
2 3 89
3 1 91
1 2 32

5 7
1 2 5
2 3 7
2 4 8
4 5 11
3 5 10
1 5 6
4 2 12

0

样例输出:

0
17
16
26

解题代码:

//* @author: ccQ.SuperSupper
import java.io.*;
import java.util.*;
class Edge implements Comparable{
	int start,end;
	int cost;
	public int compareTo(Object temp){
		Edge cnt = (Edge) temp;
		if(cnt.cost< this.cost) return 1;
		return -1;
	}
}
class Set{
	final int N = 100;
	int father[] = new int[N],num[] = new int[N];
	void init(){
		for(int i=0;i< N;++i){
			father[i] = -1;
			num[i] = 0;
		}
	}
	int find(int who){
		int u,v = who;
		while(this.father[who]!=-1){
			who = this.father[who];
		}
		while(v!=who){
			u = this.father[v];
			this.father[v] = who;
			v = u;
		}
		return who;
	}
	void set(int a,int b){
		a = this.find(a);
		b = this.find(b);
		if(a==b)
			return ;
		if(this.num[a]>this.num[b]){
			this.father[b] = a;
			this.num[a]+=this.num[b];
		}else{
			this.father[a] = b;
			this.num[b]+=this.num[a];
		}
	}
}
public class Main {
	static final int N = 100000;
	static int n,m;
	static Set set = new Set();
	static Edge edge[] = new Edge[N];
	static void start(){
		for(int i=0;i< N;++i)
			edge[i] = new Edge();
	}
	static int solve(){
		int ans=0;
		int i;
		Arrays.sort(edge,0,m);
		for(i=0;i< m;++i){
			if(set.find(edge[i].start)!=set.find(edge[i].end))
				ans+=edge[i].cost;
			set.set(edge[i].start, edge[i].end);
		}
		return ans;
	}
public static void main(String[]args) throws Exception{
 int i,j;
 start();
 StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
 while(true){
	n = Get_Num(cin);
	if(n==0) break;
	m = Get_Num(cin);
	set.init();
	for(i=0;i< m;++i){
		edge[i].start = Get_Num(cin);
		edge[i].end = Get_Num(cin);
		edge[i].cost = Get_Num(cin);
	}
	System.out.println(solve());
  }
}
	static int Get_Num(StreamTokenizer cin)throws Exception{
		cin.nextToken();
		return (int)cin.nval;
	}
}

  1. 这道题目虽然简单,但是小编做的很到位,应该会给很多人启发吧!对于面试当中不给开辟额外空间的问题不是绝对的,实际上至少是允许少数变量存在的。之前遇到相似的问题也是恍然大悟,今天看到小编这篇文章相见恨晚。

  2. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。