最小支撑树prim)

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《数据结构与算法分析》
//2013/07/01 prim#include<iostream>using namespace std;const int INIT=30000;typedef int ** Graph;void InitGraph(Graph &G,int n,int * &visted,int *&D,int *&V){int i;int j;G=new int *[n];for(i=0;i<n;++i)G[i]=new int [n];for(i=0;i<n;++i)for(j=0;j<n;++j)G[i][j]=INIT;for(i=0;i<n;++i){visted[i]=0;//访问标志D[i]=INIT;//与相连节点的距离V[i]=0;//存放相邻节点}}void CreateGraph(Graph &G,int n,int edge){   int i;   int j;   int weight;   int k;   cout<<"Please Create The Graph"<<endl;   for(k=0;k<edge;++k)   {   cin>>i>>j>>weight;   G[i][j]=G[j][i]=weight;   }}int mixvertex(Graph &G,int n,int *&D,int *&visted)//查找G中没有被访问且与相邻节点距离最小{    int i,v;for(i=0;i<n;++i)if(!visted[i] ){v=i;break;};for(i=i+1;i<n;++i){        if(!visted[i] && D[i]<D[v])v=i;} return v;}void prim(Graph &G,int s,int n,int * &visted,int *&D,int *&V){int i,v,w;V[0]=s;for(i=0;i<n;++i)//初始化第一个节点以及更新与其相邻节点的距离D[i]=G[s][i];visted[s]=1;for(i=1;i<n;++i){       v=mixvertex(G,n,D,visted);//求出       V[i]=v;   visted[v]=1;   for(w=0;w<n;++w)   {         if(!visted[w]&&D[w]>G[v][w])//更新剩余节点 { D[w]=G[v][w]; }   }}   }void print(int *&V,int n){int i;for(i=0;i<n;++i)    cout<<V[i]<<endl;}int main(){int *visted;int *D;int *V;int n=6;int edge=8;int i=0;visted=new int[n];D=new int [n];V=new int [n];Graph G;        InitGraph(G,n,visted,D,V);CreateGraph(G,n,edge);prim(G,0,n,visted,D,V);print(V,n);}