2013
11-09

# Agri-Net

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.

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.

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.

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


28

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

public class Main {

public static void main(String[] args) throws IOException {
System.in));
String s = null;
int n;
int[][] a;
int t;
String[] ss;
int index;
if (s.equals("")) {
break;
}
n = Integer.parseInt(s);
a = new int[n][n];
t = n;
while (t-- > 0) {
index = 0;
while (index != n) {
for (int i = 0; i < ss.length; i++, index++) {
a[n - t - 1][index] = Integer.parseInt(ss[i]);
}
}
}
System.out.println(prim(a, n));
}
}

public static int prim(int[][] a, int count) {
int sum = 0;
int i, j, k;
int[] lowcost = new int[count];
int[] closeset = new int[count];
boolean[] used = new boolean[count];
for (i = 0; i < count; i++) {
lowcost[i] = a[0][i];
closeset[i] = 0;
used[i] = false;
}
used[0] = true;
for (i = 1; i < count; i++) {
j = 0;
while (used[j]) {
j++;
}
for (k = 0; k < count; k++) {
if ((!used[k]) && (lowcost[k] < lowcost[j])) {
j = k;
}
}
sum += lowcost[j];
used[j] = true;
for (k = 0; k < count; k++) {
if (!used[k] && (a[j][k] < lowcost[k])) {
{
lowcost[k] = a[j][k];
closeset[k] = j;
}
}
}
}
return sum;
}
}

1. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？

3. Thanks for taking the time to examine this, I really feel strongly about it and love studying a lot more on this topic. If possible, as you acquire experience