Poj1258_Agri-Net(最小生成树)

一、Description(poj1258)

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.

Input

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity 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.

Output

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.
二、问题分析
        典型的求最小生成树问题，算法大多用Prim或者Kruskal算法。小弟用了Prim算法，这道题如果掌握了Prim算法，那么问题就迎刃而解了。可惜小弟数据结构学得不怎么踏实，有临时学了一遍，有点蛋疼。

       从单一顶点开始，普里姆算法按照以下步骤逐步扩大树中所含顶点的数目，直到遍及连通图的所有顶点。

1. 输入：一个加权连通图，其中顶点集合为V，边集合为E；
2. 初始化：V_new = {x}，其中x为集合V中的任一节点（起始点），E_new = {}；
3. 重复下列操作，直到V_new = V：
   1. 在集合E中选取权值最小的边（u, v），其中u为集合V_new中的元素，而v则不是（如果存在有多条满足前述条件即具有相同权值的边，则可任意选取其中之一）；
   2. 将v加入集合V_new中，将（u, v）加入集合E_new中；
4. 输出：使用集合V_new和E_new来描述所得到的最小生成树。

        这里我们改一下算法，是输出的是权值和。

三、Java代码

import java.util.Scanner;

public class Test {
	public static int prim(int[][] a, int count) {
        int sum = 0;
        int i, j, k;
        int[] lowcost = new int[count]; //保存相关定点间的权值
        int[] adjvex = new int[count]; //保存相关定点的下标
        for (i = 0; i < count; i++) {
            lowcost[i] = a[0][i];
            adjvex[i]=0;
        }
        for (i = 1; i < count; i++) {
            j=0;
            while(lowcost[j]==0){
            	j++;
            }
            for (k = 1; k < count; k++) {
                if ((lowcost[k]!=0) && (lowcost[k] < lowcost[j])) {
                    j = k;
                }
            }
            sum += lowcost[j];
            lowcost[j] = 0;
            for (k = 1; k < count; k++) {
                if (lowcost[k]!=0 && (a[j][k] < lowcost[k])) {
                    {
                        lowcost[k] = a[j][k];
                        adjvex[k] = j;
                    }
                }
            }
        }
        return sum;
    } 
	public static void main(String[] args) {
		 
		  Scanner cin=new Scanner(System.in);
		  int n;
		  while((n=cin.nextInt())!=-1){
		  int vertexNumber = n;
	      int arr[][] = new int[vertexNumber][vertexNumber];
	      for (int i = 0; i < vertexNumber; i++) {  
	          for (int j = 0; j < vertexNumber; j++) {  
	              arr[i][j]=cin.nextInt();
	          }  
	      }  
	      System.out.println(prim(arr,n));
	      n=-1;
		  }
	}
}

版权声明：本文为博主原创文章，未经博主允许不得转载。