poj 1258 Agri-Net-----最小生成树 prim

最小生成树算法在农网互联中的应用

最新推荐文章于 2025-11-27 20:48:17 发布

最新推荐文章于 2025-11-27 20:48:17 发布 · 94 阅读

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#c/c++

本文探讨了最小生成树算法在农民互联网连接项目中的应用，旨在通过最小化光纤使用量来降低成本，实现所有农场之间的高效互联。

Agri-Net

Time Limit:1000MS		Memory Limit:10000K
Total Submissions:26418		Accepted:10425

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

题目大意：将已知距离的一些点连接起来，总的距离最少。

思路：以一个点为起始，每次找下一个最短边，直到找到最小生成树。

//Memory: 208 KB		Time: 16 MS
//Language: C++		Result: Accepted

#include<stdio.h>
#include<string.h>
const int MAX=1000050;
int dis[105][105],n,vir[105],t[105],sum;
void minDis(int n);
int main()
{
	//freopen("in.txt","r",stdin);
	int i,j;
	while(scanf("%d",&n)!=EOF)
	{
		memset(vir,0,sizeof(vir));
		sum=0;
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				scanf("%d",&dis[i][j]);
		minDis(n);
		printf("%d\n",sum);
	}
	return 0;
}

void minDis(int n)
{
	int i,j,temp=n,next=0,min,count=0;
	vir[0]=1;
	while(--temp)
	{
		min=MAX;
		t[count++]=next;             //把新找到的一个最小树的点加入t数组
		for(j=0;j<count;j++)
		{
			for(i=0;i<n;i++)
			{
				if(dis[t[j]][i]<min && !vir[i])
				{
					min=dis[t[j]][i];
					next=i;
				}
			}
		}
		vir[next]=1;        //标记
		sum+=min;
		//printf("%d %d %d\n", min, count ,next);
	}
}