E - More is better

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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. 

InputThe 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) OutputThe 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

#include
#include
#include
#include
using namespace std;
#define maxn 100010
int par[maxn],num[maxn],sum[maxn];
void init()
{
	for(int i=1;i<=maxn;i++)
	{
	par[i]=i;
	num[i]=1;
    }
}
int find(int x)
{ if(x!=par[x])
 par[x]=find(par[x]);  
    return par[x];  
}
void unite(int a,int b)
{
	int fa=find(a);
	int fb=find(b);
	if(fa!=fb)
    {
	par[fa]=fb;
  }
}
int main()
{int n,a,b;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
printf("%d\n",1);
else{
	init();
	int max=0;
	while(n--)
	{
	scanf("%d%d",&a,&b);
	if(a>max)
	max=a;
	if(b>max)
	max=b;
	unite(a,b);
   }
   memset(sum,0,sizeof(sum));
   for(int i=1;i<=max;i++)
   sum[find(i)]++; 
int max1=0;
for(int i=1;i<=max;i++)
{
	if(sum[find(i)]>max1)
	max1=sum[find(i)];
}
printf("%d\n",max1);
}
}
	return 0;
}

41 23 45 67 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.
 
        
 

分析：平时有时候会让求集合（畅通工程题），而本题用并查集求得每个集合的元素的个数，最后输出最大的集合的元素个数；

代码：

#include
#include
#include
#include
using namespace std;
#define maxn 100010
int par[maxn],num[maxn],sum[maxn];
void init()
{
	for(int i=1;i<=maxn;i++)
	{
	par[i]=i;
	num[i]=1;
    }
}
int find(int x)
{ if(x!=par[x])
 par[x]=find(par[x]);  
    return par[x];  
}
void unite(int a,int b)
{
	int fa=find(a);
	int fb=find(b);
	if(fa!=fb)
    {
	par[fa]=fb;
  }
}
int main()
{int n,a,b;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
printf("%d\n",1);
else{
	init();
	int max=0;
	while(n--)
	{
	scanf("%d%d",&a,&b);
	if(a>max)
	max=a;
	if(b>max)
	max=b;
	unite(a,b);
   }
   memset(sum,0,sizeof(sum));
   for(int i=1;i<=max;i++)//统计各个集合的个数
   sum[find(i)]++; 
int max1=0;
for(int i=1;i<=max;i++)
{
	if(sum[find(i)]>max1)
	max1=sum[find(i)];
}
printf("%d\n",max1);
}
}
	return 0;
}