Solving the minimum spanning tree greedy algorithm

First, the object of the experimental requirements

1. Familiar with the greedy algorithm of the basic principles and scope of application.
2. Use greedy algorithm programming, solving the minimum spanning tree problem.

Second, the experimental content

Optionally one greedy algorithm (Prim or Kruskal), solving the minimum spanning tree. Description of the algorithm and the complexity of the analysis.

Programming, and provide the test example

Third, to achieve ideological

Prim: S, V into two sets, as the initial S {1}, V contains all vertices, and then select the nearest vertex distance from the vertex S of the VS j, and j are added to the S. Setting two arrays closest and lowcost. For each j belongs VS, closet [j] is adjacent to the point j in S, in which S k other neighbor is compared with a v [j] [closet [j ]] <= v [j] [k] wherein lowcost [j] is the value of V [J] [Closest] . Is the corresponding weight value of the last minimum spanning tree lowcost stored, and for each vertex, Closest minimum spanning tree can be found in the corresponding edge.

Fourth, the implementation code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define mmax 100
#define maxt 100000000
int v[mmax][mmax];

void Prim(int n){
	int low[mmax];// 
	int clo[mmax];
	bool s[mmax];//判断该节点是否在S中 
	s[1]=true;
	for(int i=2;i<=n;i++){
		low[i]=v[1][i];//初始化 
		clo[i]=1;
		s[i]=false;
	}
	for(int i=1;i<n;i++){//对剩余的n-1个顶点进行操作 ,此处的i并无实际意思，只是起到一个记录变量的作用 
		int min=maxt;
		int j=1;
		for(int k=2;k<=n;k++)//从V-S中找到S中边权最小的节点 
			if((low[k]<min)&&(!s[k])){
				min=low[k];
				j=k;
			}
			cout<<j<<" "<<clo[j]<<":"<<low[j]<<endl;
			s[j]=true;
			for(int k=2;k<=n;k++){//调整low中的值，加入j节点之后，使得两个集合中的距离最小 
				if((v[j][k]<low[k])&&(!s[k])){
					low[k]=v[j][k];
					clo[k]=j;
				}
			}
		
	}
} 

int main()
{
	int n,m;//n vertexs and m edges
	cin>>n>>m;
	int ii,jj,tt;
	for(int i=0;i<=n;i++){
		for(int j=0;j<=n;j++)
		v[i][j]=maxt;
	}
	for(int i=0;i<m;i++){
		cin>>ii>>jj>>tt;
		v[ii][jj]=tt;
		v[jj][ii]=tt;
	}
	Prim(n);
	return 0;
}

//6 10
//1 2 6
//1 4 5
//1 3 1
//2 3 5
//2 5 3
//3 4 5
//3 5 6
//3 6 4
//4 6 2
//5 6 6