Prim算法---最小生成树

#include<iostream>  
#include<cstdio>  
#include<map>  
#include<cstring>  
#include<string>  
#include<algorithm>  
#include<queue>  
#include<vector>  
#include<stack>  
#include<cstdlib>  
#include<cctype>  
#include<cstring>  
#include<cmath>  
using namespace std;  
const int inf=0x3f3f3f3f;  
int G[1001][1001];//邻接矩阵  
int vis[1001],lowc[1001];  
int prim(int G[][1001],int n){  
    int i,j,p,minc,res=0;  
    memset(vis,0,sizeof(vis));  
    vis[1]=1;//从1开始访问  
    for(i=2;i<=n;i++) lowc[i]=G[1][i];  
    for(i=2;i<=n;i++){  
        minc=inf; p=-1;  
        for(j=1;j<=n;j++){  
            if(vis[j]==0&&lowc[j]<minc){  
                minc=lowc[j]; p=j;  
            }  
        }  
        //cout<<minc<<endl;  
        if(inf==minc) return -1;//原图不联通  
        res+=minc; vis[p]=1;  
        for(j=1;j<=n;j++){//更新lowc[]  
            if(vis[j]==0&&lowc[j]>G[p][j]){  
                lowc[j]=G[p][j];  
            }  
        }  
    }  
    return res;  
}  
  
int main()  
{  
    int n,m;  
    while(cin>>n>>m){  
        int x,y,w;  
        memset(G,inf,sizeof(G));//首先记录所有边的权为inf  
        for(int i=1;i<=m;i++){  
            cin>>x>>y>>w;  
            G[x][y]=G[y][x]=w;  
            //cout<<G[x][y]<<endl;  
        }  
        //int res=prim(n);  
        cout<<prim(G,n)<<endl;  
    }  
    return 0;  
}  

#include<stdio.h>
#include<string.h>
#include<algorithm>

using namespace std;

struct tr
{
	int s,e,w;
}p[50000+10];

bool cmp(tr x, tr y)
{
	return x.w < y.w;
}

int n,m;
int f[1000+10];
int i,j;
long long ans;

int find(int x)  //找父亲 
{  
	int r = x;
	while(f[r] != r)
	r = f[r];
	return r;
	int i = x, j;
	while(i != r)
	{
		j = f[i];
		f[i] = r;
		r = j;
	}   
} 

void join(int x, int y)
{
	int fx = find(x);
	int fy = find(y);
	if(fx != fy)
		f[fx] = fy;
} 

int kruskal()
{
	sort(p, p+m, cmp);
	for(i=0; i<n; i++)
	{
		f[i] == i;//初始化 父亲节点 
	}
	for(i=0; i<n; i++)
	{
		if(find(p[i].s) != find(p[i].e))
		{
			join(p[i].e, p[i].s);
			ans += p[i].w;
		}
	}
	return ans;
}

int main()
{
	ans=0;
	scanf("%d %d",&n,&m);
	for(i=0; i<m; i++)
	{
		scanf("%d %d %d",&p[i].s, &p[i].e, &p[i].w);
	}
	kruskal();
	printf("%lld\n",ans);
	return 0;
 }