克鲁斯卡尔（Kruskal）算法 Agri-Net

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最新推荐文章于 2021-04-30 20:56:39 发布

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分类专栏：克鲁斯卡尔算法

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本文详细介绍了克鲁斯卡尔（Kruskal）算法，并结合农业网络（Agri-Net）的应用，展示了算法如何解决实际问题。通过输入输出示例，读者可以深入理解算法的运作过程。

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克鲁斯卡尔（Kruskal）算法：先将权值从小到大排序，再用并查集思想

Agri-Net

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45184		Accepted: 18553

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

Sample Output

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct note{
	int q,z,cost;
}s[10100];
int per[110];
int sum,n,k;
bool cmp(note a1,note a2)
{
	return a1.cost<a2.cost;
}
void init()
{
	for(int i=1;i<=n;i++)
	per[i]=i;
}
int fun(int x)
{
	int r=x;
	while(per[r]!=r)
	r=per[r];
	per[x]=r;
	return r;
}
int join(int x,int y)
{
	int fx=fun(x);
	int fy=fun(y);
	if(fx!=fy)
	{
		per[fx]=fy;
		return 1;
	}
	return 0;
}
int kru()
{
	int j=0,i;
	init();
	sort(s,s+k,cmp);
	sum=0;
	for(i=0;i<k;i++)
	{
		if(join(s[i].q,s[i].z))
		{
			j++;
		sum+=s[i].cost;
		}
	    if(j==n-1)
	    return sum;
	}
	return j;
}
int main()
{
	int cost1,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		k=0;
		for(i=1;i<=n;i++)
		{
		for(j=1;j<=n;j++)	
		 {
		  scanf("%d",&cost1);
			if(i!=j){
			
			s[k].q=i;
			s[k].z=j;
			s[k++].cost=cost1;
		}
		}
		}
	   printf("%d\n",kru());
	}
	return 0;
}