算法代码如下

#include <iostream>
using namespace std;
const int INF = 0x3fffffff;
const int N = 100;
bool s[N];
int closest[N];
int lowcost[N];
void Prim(int n, int u0, int c[N][N]) 
{  //顶点个数n、开始顶点u0、带权邻接矩阵C[n][n]
  //如果s[i]=true,说明顶点i已加入最小生成树
  //的顶点集合U；否则顶点i属于集合V-U
  //将最后的相关的最小权值传递到数组lowcost
  s[u0] = true; //初始时，集合中U只有一个元素，即顶点u0
  int i;
  int j;
  for(i = 1; i <= n; i++)//①
  {
       if(i != u0) 
       {
            lowcost[i] = c[u0][i];
            closest[i] = u0;
            s[i] = false;
       }
       else
            lowcost[i] =0;
  }
  for(i = 1; i <= n; i++)  //②
  {
       int temp = INF;
       int t = u0;
       for(j = 1; j <= n; j++) //③在集合中V-u中寻找距离集合U最近的顶点t
       { 
           if((!s[j]) && (lowcost[j] < temp)) 
           {
                t = j;
                temp = lowcost[j];
           }
       }
      if(t == u0)
         break;       //找不到t，跳出循环
      s[t] = true;    //否则，讲t加入集合U
      for(j = 1; j <= n; j++)  //④更新lowcost和closest
      {       
          if((!s[j]) && (c[t][j] < lowcost[j]))
          {
              lowcost[j] = c[t][j];
              closest[j] = t;
          }
      }
    }
}
int main()
{
    int n, c[N][N], m, u, v, w;
    int u0;
    cout <<"输入结点数n和边数m："<<endl;
    cin >> n >> m;
    int sumcost = 0;
    for(int i = 1; i <= n; i++) 
       for(int j = 1; j <= n; j++) 
          c[i][j] = INF;
    cout <<"输入结点数u，v和边值w："<<endl;
    for(int i=1; i<=m; i++) 
    {
        cin >> u >> v >> w;
        c[u][v] = c[v][u] = w;
    }
    cout <<"输入任一结点u0："<<endl;
    cin >> u0 ;
    //计算最后的lowcost的总和，即为最后要求的最小的费用之和
    Prim(n, u0, c);
    cout <<"数组lowcost的内容为："<<endl;
    for(int i = 1; i <= n; i++)
        cout << lowcost[i] << " ";
    cout << endl;
    for(int i = 1; i <= n; i++)
                sumcost += lowcost[i];
    cout << "最小的花费是：" << sumcost << endl << endl;
    return 0;
}