Agri-Net Prim（Kruskal）的应用

最新推荐文章于 2022-05-25 18:11:02 发布

原创最新推荐文章于 2022-05-25 18:11:02 发布 · 334 阅读

0 ·

CC 4.0 BY-SA版权

图论之搜索问题模板专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一种高效的算法，用于解决如何用最少的光纤连接多个农场的问题。通过构建最小生成树的方法，确保所有农场都能以最低成本连接到互联网。该算法首先读取农场之间的距离矩阵，然后使用贪心算法挑选最短路径进行连接。

描述

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.

Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.

Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.

The distance between any two farms will not exceed 100,000.

输入

INPUT FORMAT For Each Test Data

Line 1:	The number of farms, N (3 <= N <= 100).
Line 2...N+1:	The subsequent lines contain the N x N connectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

输出

OUTPUT FORMAT For Each Test Data

The single output contains the integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

样例输入

4
0 4 9 21
4 0 8 17
9 8 0 16
21 17 16 0

样例输出

代码如下：

import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

class map implements Comparable {
	int length;
	int x;
	int y;
	public int compareTo(Object o) {
		map s = (map) o;
		return length - s.length;
	}
}

public class Main {
    ///一个标记的数组
	private static int flag[];
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int nodenumber = sc.nextInt();
		List<map> list = new ArrayList<map>();
		for (int i = 0; i < nodenumber; i++) {
			for (int j = 0; j < nodenumber; j++) {
				map data = new map();
				data.length = sc.nextInt();
				data.x = i;
				data.y = j;
				list.add(data);
			}
		}
		/// 开始每一个节点的父节点，初始化为自己。
		flag = new int[nodenumber];
		for (int i = 0; i < nodenumber; i++)
			flag[i] = i;
		/// 开始选择，节点，建造路径。
         Collections.sort(list);
         ///初始化一些值
         int sum =0 ,edge = 0;
         for(int i=0;i<list.size();i++)
         {
        	 if(!union(list.get(i).x,list.get(i).y))
        		 continue;
        	 edge++;
        	 sum+= list.get(i).length;
        	 if(edge==nodenumber-1)
        		 break;
         }
         System.out.println(sum);
	}
	
	private static int find(int p)  
	{   
	    while (p != flag[p]) 
	    	p = flag[p]; 
	    return p;  
	}  
	public static boolean union(int p, int q)  
	{   
	    int pRoot = find(p);  
	    int qRoot = find(q);  
	    ///一个村子里面的，就不需要搬
	    if (pRoot == qRoot)   
	        return  false;  
	    ///个人喜好问题，是搬到哪边村子，同名
	    flag[pRoot] = qRoot; 
	   return true;
	}  
}