最小生成二叉树-prim算法

1.prim算法：一种计算生成最小生成树的方法，它的每一步都会为一棵生长中的树添加一条边.

2.时间复杂度：

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通过邻接矩阵图表示的简易实现中，找到所有最小权边共需O（V）的运行时间。使用简单的二叉堆与邻接表来表示的话，普里姆算法的运行时间则可缩减为 O(ElogV)，其中E为连通图的边数，V为顶点数。如果使用较为复杂的斐波那契堆，则可将运行时间进一步缩短为O(E+VlogV)，这在连通图足够 密集时（当E满足Ω（VlogV）条件时），可较显著地提高运行速度。

3.代码实现：

 #include <stdio.h>
 #define    MAXV 100  //最大顶点个数
 #define INF 32767 //INF表示∞
 typedef struct
 {      int edges[MAXV][MAXV];//邻接矩阵
        int vexnum,arcnum;   //顶点数，弧数
 } MGraph;//图的邻接矩阵类型

 void init(MGraph &g);//初始化邻接矩阵
 void DispMat(MGraph g);//输出邻接矩阵g
 void prim(MGraph g,int v);
 int main()
 {
     int u=3;
     MGraph g;//图的邻接矩阵
     init(g);//初始化邻接矩阵
     printf("图G的邻接矩阵:\n");
     DispMat(g);
     printf("\n");
     printf("普里姆算法求解结果:\n");
     prim(g,0);
     printf("\n");
     return 0;
 }

 void prim(MGraph g,int v)//从v号节点开始---生成最小生成树
 {
     //(V-U)---未加入最小生成树的点
     //U---已加入最小生成树的点
     int i,j,k;
     int MinCost[MAXV];   //(V-U)中各点离U的最小距离
     int MinCostNum[MAXV];//(V-U)中各点离U的最小距离对应在U中的点
     int min;//min记录离U最近的距离
     MinCost[v]=0;//v加入U
     for (i=0;i<g.vexnum;i++) //初始化MinCost[]和MinCostNum[]
     {
         MinCost[i]=g.edges[v][i];//每个节点距v的值
         MinCostNum[i]=v;//(V-U)中的节点i距U中最近的点是v
     }
     for (i=1;i<g.vexnum;i++)
     {
         min=INF;
         for (j=0;j<g.vexnum;j++)//在(V-U)中找出离U最近的顶点k
                if (MinCost[j]!=0 && MinCost[j]<min) //未加入U(即V-U)中的点且距离U最近
             {
                 min=MinCost[j];
                 k=j; //k记录离U最近的顶点
             }
             if(min!=INF)//(V-U)中离U最近点k--距U中最近的一个点是MinCostNum[k]
                 printf("边(%d,%d)权为:%d\n",MinCostNum[k],k,min);
             MinCost[k]=0;//标记k已经加入U
             //更新(V-U)中的点/////////////////////
             for (j=0;j<g.vexnum;j++)//由于顶点k的新加入而修改数组lowcost和closest
                    if (MinCost[j]!=0 && g.edges[k][j]<MinCost[j])
                 {
                     MinCost[j]=g.edges[k][j];
                     MinCostNum[j]=k;
                 }
             //经典之处///////////////////////////////////////
     }
 }
 void init(MGraph &g)
 {
     int i,j;
     g.vexnum=6;g.arcnum=10;
     int A[MAXV][11];
     for (i=0;i<g.vexnum;i++)
         for (j=0;j<g.vexnum;j++)
             A[i][j]=INF;
     //数据结构书P189---图7.34
     A[0][2]=1;A[0][4]=3;A[0][5]=7;
     A[1][2]=9;
     A[2][3]=2;
     A[3][5]=4;
     A[4][3]=6;A[4][5]=8;
     /*for (i=0;i<g.vexnum;i++)//使邻接矩阵对称
         for (j=0;j<g.vexnum;j++)
             A[j][i]=A[i][j];*/
     for (i=0;i<g.vexnum;i++)//建立邻接矩阵
         for (j=0;j<g.vexnum;j++)
             g.edges[i][j]=A[i][j];

 }
 void DispMat(MGraph g)//输出邻接矩阵g
 {
     int i,j;
     for (i=0;i<g.vexnum;i++)
     {
         for (j=0;j<g.vexnum;j++)
             if (g.edges[i][j]==INF)
                 printf("%3s","∞");
             else
                 printf("%3d",g.edges[i][j]);
         printf("\n");
     }
 }
/*
图G的邻接矩阵:
 ∞ ∞ 1 ∞ 3 5
 ∞ ∞  7 ∞ ∞ ∞
 ∞ ∞ ∞ 9 ∞ ∞
 ∞ ∞ ∞ ∞ ∞ 2
 ∞ ∞ ∞ 4 ∞ 6
 ∞ ∞ ∞ ∞ ∞ ∞
普里姆算法求解结果:
边(0,2)权为:1
边(0,4)权为:3
边(4,3)权为:6
边(3,5)权为:4
*/