ZOJ 1542//POJ1861这题给的样例有点坑爹啊！木有按照样例输出，居然过了。啥破题啊！样例与实际的题意不符，这是误导了好多孩纸啊！

题目的本意应该是叫你求出最小生成树的最大边，和边的个数，最后把每条边的连接情况打印出来。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define N 15001

int father[1001];//按照模板敲的
int  p[N], n, m;
typedef struct 
{
   int x, y;
   int w; 
}edge;
edge e[N];

int cmp(const void *a,const void *b)
{
    return (*(edge *)a).w > (*(edge *)b).w ? 1 : -1;
}   
    
int Make_set(int x)
{
    father[x] = x;    
}    
    
int Find_set(int x)
{
    if(x != father[x] )
      father[x] = Find_set(father[x]);    
    return father[x];
} 
 
void Kruskal()
{    
   int i,a,b,count,max;  
   qsort(e+1,m,sizeof(edge),cmp);  
   for(i=1; i<=n; i++)
    Make_set(i);
   max = -1; count = 0;
   for(i=1; i<=m; i++)  
   {  
      a = Find_set(e[i].x); 
      b = Find_set(e[i].y);
      if(a != b)  
      {
        p[++count] = i;  
        father[a] = b;
        if(e[i].w > max)
          max = e[i].w;
      }
    } 
   printf("%d\n%d\n",max,count);  
   for(i=1; i<=count; i++)
   {
      printf("%d %d\n",e[ p[i] ].x,e[ p[i] ].y);       
   } 
}     

int main()
{
   int i;
     
  while(scanf("%d%d", &n, &m) !=  EOF)
  { 
     for(i=1; i<=m; i++)
     { 
       scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
     } 
     Kruskal(); 
   }    
  // system("pause");
   return 0;
}