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
*/