1.什么是最小生成树算法？

2.Prim算法的步骤是什么？

a.假定图的顶点集合为V，边集合为E.
b.初始化点集合U={u}.//u为V中的任意选定的一点
c.从u的邻接结点中选取一点v使这两点之间的权重最小,然后将v加入集合U中.
d.从结点v出发，重复c步骤，直到V={}.

3.举个例子来说明Prim算法的步骤：

4.Prim算法的具体C语言编程实现：

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 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374 #include#include#includeconst int Max =0x7fffffff;const int N=50;  int n;int g[N][N],dis[N],visited[N];  int prim(){  int i,j;  int pos,min;  int ans=0;  memset(visited,0,sizeof(visited));  visited[1]=1;pos=1;  //assign a value to the dis[N] first  for(i=2;i<=n;i++)    dis[i]=g[pos][i];  for(i=1;idis[j])      {        min=dis[j];        pos=j;       }    }    printf("The node being traversed is :%d\n",pos);    ans+=min;    printf("The value of ans is %d\n",ans);    //mark the node    visited[pos]=1;    //update the weight    for(j=1;j<=n;j++)      if(visited[j]==0&&dis[j]>g[pos][j])        dis[j]=g[pos][j];  }  return ans;}  int main(){  int i=1,j=1;  int ans=0;  int w;  printf("Please enter the number of the nodes:\n");  scanf("%d",&n);  for(i=1;i<=n;i++)    for(j=1;j<=n;j++)    {      if(i==j)        g[i][j]=0;      else        g[i][j]=Max;    }  printf("Please enter the number of the edges:\n");  int edgenum;  scanf("%d",&edgenum);  int v1,v2;  printf("Please enter the number and the corresponding weight:\n");  for(i=1;i<=edgenum;i++)  {    scanf("%d%d%d",&v1,&v2,&w);    g[v1][v2]=g[v2][v1]=w;  }  ans=prim();  printf("The sum of the weight of the edges is:%d\n",ans);  system("pause");  return 0;    }

5.程序运行后的结果截图