Agri_Net

题目：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.
思路：最小生成树的prim算法

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Agri_Net1 {

    public static void main(String[] args) throws IOException { 
        BufferedReader read = new BufferedReader(new InputStreamReader( System.in)); 
        String s = null; 
        int n; 
        int[][] a; 
        int t; 
        String[] str; 
        int index; 
        while ((s = read.readLine()) != null) { 
            if (s.equals("")) { break; } 
            n = Integer.parseInt(s); 
            a = new int[n][n]; 
            t = n; 
            while (t--> 0) {
                 index=0; 
                 while (index!=n) { 
                 str = read.readLine().split(" "); 
                 for (int i=0; i<str.length;i++, index++) { 
                    a[n-t-1][index] = Integer.parseInt(str[i]); 
                 }
             } 
        } 
        System.out.println(prim(n, a)); 
        } 
    } 
    public static int prim(int count,int [][]arr){
        int sum = 0; 
        int i, j, k; 
        int[] lower = new int[count]; 
        int[] edge = new int[count]; 
        boolean[] checked = new boolean[count]; 
        for (i = 0; i < count; i++) { 
            lower[i] = arr[0][i]; 
            edge[i] = 0; 
            checked[i] = false; 
        } 
        checked[0] = true; 
        for (i = 1; i < count; i++) { 
            j = 0;
            while (checked[j]) { j++; } 
            for (k = 0; k < count; k++) { 
                if ((!checked[k]) && (lower[k] < lower[j])) { j = k; } 
            } 
            sum += lower[j]; 
            checked[j] = true; 
            for (k = 0; k < count; k++) { 
                if (!checked[k] && (arr[j][k] < lower[k])) { 
                { 
                lower[k] = arr[j][k]; 
                edge[k] = j;
                } 
                } 
            }
        } 
        return sum;     
    }
}