Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37109		Accepted: 14982

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

#include<cstdio>

#include<iostream>

#include<cmath>

#include<algorithm>

#include<cstring>

#define MAX 600*300

using namespace std;

struct Node

{

    int a,b,c;

};

Node node[MAX];

int parent[MAX];

int num[][];

int sum;

int temp;

int Find(int x)

{

    return x==parent[x]?x:parent[x]=Find(parent[x]);

}

void Kruskal()

{

    int x,y;

    int i,j;

    for(i=;i<temp;i++)

    {

        x=node[i].a;

        y=node[i].b;

        x=Find(x);

        y=Find(y);

        if(x!=y)

        {

            parent[x]=y;

            sum+=node[i].c;

        }

    }

}

bool cmp(Node a,Node b)

{

    return a.c<b.c;

}

int main()

{

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        sum=;

        int i,j;

        for(i=;i<MAX;i++)

            parent[i]=i;

        for(i=;i<n;i++)

        {

            for(j=;j<n;j++)

            {

                   scanf("%d",&num[i][j]);

            }

        }

        temp=;

        for(i=;i<n;i++)

        {

            for(j=;j<n;j++)

            {

                node[++temp].a=i;

                node[temp].b=j;

                node[temp].c=num[i][j];

            }

        }

        sort(node,node+temp,cmp);

        Kruskal();

        printf("%d\n",sum);

    }

    return ;

}