poj 1861 network （kruskal）

题意：Andrew 打算建一个 a new network in his company. 要求所用电缆最短，给出n个hubs
m条能够相连的边，求出最小生成树中最大的边，且输出所有相连的边

思路：典型的kruskal；

#include "stdio.h"

#include <algorithm>

#include <string.h>

#define M 15005

using namespace std;

struct data

{

int
v1,v2,w;

}edge[M];

bool cmp (const data &a,const data
&b) //比较函数必须为
const 防止数据改变 且最好用&（引用而不用指针）

{

return a.w
< b.w;

}

int find (int father[],int v)

{

int t;

t = v;

while
(father[t] >= 0)

t = father[t];

return
t;

}

void kruskal (int n,int m)

{

int
father[M],vf1,vf2,i,j;

data
T[M];

memset
(father,-1,sizeof(father));

i = 0;

j = 0;

while (i
< m&&j
< n-1)

{

vf1 = find (father,edge[i].v1);

vf2 = find (father,edge[i].v2);

if (vf1 != vf2)

{

father[vf2] = vf1;

T[j++] =
edge[i];
//把边加入T中

}

i ++;

}

int max =
0;

for (i = 0;i
< n-1;i ++)

if (max < T[i].w)

max = T[i].w;

printf
("%d\n%d\n",max,n-1);

for (i = 0;i
< n-1;i ++)

printf ("%d %d\n",T[i].v1,T[i].v2);

}

int main ()

{

int
n,m,i,x,y,cost;

scanf ("%d
%d",&n,&m);

for (i = 0;i
< m;i ++)

{

scanf ("%d %d
%d",&x,&y,&cost);

edge[i].v1 = x;

edge[i].v2 = y;

edge[i].w = cost;

}

sort
(edge,edge+m,cmp);

kruskal(n,m);

return
0;

}