HDU 1856 More is better

                             More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 720    Accepted Submission(s): 264

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

 

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

 

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

 

Sample Input

4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

 

Sample Output

4
2

Hint

A and B are friends(direct or indirect), B and C are friends(direct or indirect), then A and C are also friends(indirect). In the first sample {1,2,5,6} is the result. In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

 

Author

lxlcrystal@TJU

 

Source

HDU 2007 Programming Contest - Final

 

Recommend

lcy

 

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// 1256387 2009-04-10 22:20:57 Wrong Answer 1856 250MS 78532K 804 B C++ Wpl 
// 1256393 2009-04-10 22:21:44 Accepted 1856 250MS 78532K 880 B C++ Wpl 
// 改变的并查集,实际上就是kruskal求最小生成树的思想
#include  < iostream >
#define  MAX 10000002   // 因为这个题目内存限制是102400 K ,所以开了这么大的数组
using   namespace  std;
struct  node
{
  int  parent;
  int  w;
}data[MAX];
int  MM;
void  Init()
{
  int  i;
  for (i = 1 ;i < MAX;i ++ )
 {
  data[i].parent =- 1 ;   // 这里利用的是kruskal的思想,这里父亲结点全初始化为-1
  data[i].w = 1 ;
 }
}
int  Find( int  x)   // 这里是用来查找根的,就是当前这个x的根节点
{
  int  i = x;
  while (data[i].parent > 0 )
  i = data[i].parent;
  return  i;
}
void  Merge( int  a, int  b)
{
  if (data[a].w > data[b].w)
 {
  data[a].w += data[b].w;
  data[b].parent = a;
   if (data[a].w > MM)   // 这里用来统计哪个集合的元素个数最多
   MM = data[a].w;
 }
  else
 {
  data[b].w += data[a].w;
  data[a].parent = b;
   if (data[b].w > MM)
   MM = data[b].w;
 }
}
int  main()
{
  int  n;
  while (scanf( " %d " , & n) != EOF)
 {
  MM = 1 ;
   int  a,b,x,y;
  Init();
   while (n -- )
  {
   scanf( " %d%d " , & a, & b);
   x = Find(a);
   y = Find(b);
    if (x != y)
    Merge(x,y);
  }
  printf( " %d\n " ,MM);
 }
  return   0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/11/1454499.html