poj1258Agri-Net（最小生成树）

题目链接：http://poj.org/problem?id=1258

Description

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.

Sample Input

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

Sample Output

28

对于最小生成树，一般而言会有两种通用算法：
一：Kruskal算法：对边进行贪心操作（每条边一般而言会有三种属性值：val:边的权值、(x,y):边连接的两个点的编号）
　　1）将所有的边按照边的权值进行从小到大的排序（结构体类型的排序，三个属性值是一体的）；
　　2）设一棵树为T，将边从小到大进行遍历，若是加入这条边不会形成圈（形成圈了，则一定不是树，便一定权值和不是最小），就将这条边加入树T中，如此循环，
　　　 直到遍历完所有的边。最后便会生成一颗最小生成树。（至于如何判断假如一条边会不会形成圈，可以使用并查集的知识进行实现（若新加入的边的两个顶点
　　　 不在一个集合中，则不会形成圈））。
poj1258的Kruskal的AC代码：

#include<iostream> //利用最小生成树中的kruskal算法
#include<cstdio>
#include<algorithm>
using namespace std;
int bin[120];
struct node{
int val,x,y;
}a[10010];
int cmp(struct node a,struct node b){
return a.val<b.val;
}
int find(int x){
if(bin[x]==x) return x;
else return bin[x]=find(bin[x]);
}
void unite(int x,int y){
x=find(x),y=find(y);
bin[x]=y;
}
int main(){
int n,d;
struct node c;
while(cin>>n){
int k=0;
for(int i=1;i<=n;i++) //利用并查集的知识防止生成圈，若生成圈则一定不是最小
bin[i]=i;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%d",&d);
if(j>i) a[k].val=d,a[k].x=i,a[k].y=j,k++;
}
sort(a,a+k,cmp); //对边的长度进行排序
int sum=0;
for(int i=0;i<k;i++){//对边从小到大进行遍历
if(find(bin[a[i].x])!=find(bin[a[i].y])) unite(bin[a[i].x],bin[a[i].y]),sum+=a[i].val; //若不会生成圈，则放入集合
}
printf("%d\n",sum);
}
return 0;
}