hdu 1856 关于并查集的（很简单的并查集）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 b

hdu 1856 关于并查集的（很简单的并查集）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 b
hdu 1856 关于并查集的（很简单的并查集）
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.
#include
#include
#include
using namespace std;
int a[10000002];
int b[10000002];
int find(int x)
{
if(a[x]!=x)
a[x]=find(a[x]);
return a[x];
}///////////////////////用递归可以找到根节点,还可以压缩路径
int max;
void hb(int x,int y)
{
x=find(x);
y=find(y);
if(b[x]max)////////////////////在并的过程直接找最大值,max是全局变量
max=b[y];
}
else
{
a[y]=x;
b[x]+=b[y];
if(b[x]>max)
max=b[x];
}
}//////////////////////这个是传两个点的合并过程
int main()
{
int i,j,k,m,n;
while(scanf("%d",&k)!=EOF)
{
if(k==0)
{
printf("1\n");
continue;
}
for(i=1;i

hdu 1856 关于并查集的（很简单的并查集）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 b
你没有考虑到成环的情况
比如
3
1 2
2 3
3 4
再改下 争取AC啊