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最新推荐文章于 2022-02-25 23:41:44 发布

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#树 #图 #算法 #prim

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本文介绍了一种利用Prim算法解决农场互联网连接问题的方法，通过构建最小生成树来确定连接所有农场所需的最短光纤总长度。文章提供了一个具体的示例输入和输出，并给出了完整的C++实现代码。

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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

//用prim最小生成树算法可以很容易解决

#include<iostream>
using namespace std;
void prim(int **a,int n)//一个二维数组及边长
{
   int *select=new int [n];
   memset(select,0,n*sizeof(int));
int sum=0;
int selnum=1;//已经选出的节点个数
select[0]=1;//从1开始
int smallnum;//记录最小的长度
int ir,jr;//寄存最小线段的起始点与终点
while(selnum!=n)
{smallnum=100000;
for(int i=0;i<n;i++)
{
      if(select[i]!=0)
      for(int j=0;j<n;j++)
      {if((smallnum>a[i][j])&&(select[j]==0)) {smallnum=a[i][j];ir=i;jr=j;}}
}
select[jr]=1;
sum+=smallnum;
selnum++;
}
cout<<sum<<endl;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{ int **a=new int*[n];
    for(int i=0;i<n;i++)
      a[i]=new int[n];
    for( i=0;i<n;i++)
       for(int j=0;j<n;j++)
          cin>>a[i][j];
prim(a,n);

}

return 0;
}