题目链接：http://icpc.njust.edu.cn/Problem/Hdu/1856/
题目就是要求并查集中各树的大小的最大值，我们只要在根节点处存树的大小就可以，合并也是合并根节点的数，最后扫一遍即可。
代码如下：
#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
#define pf printf
#define mem(a,b) memset(a,b,sizeof(a))
#define prime1 1e9+7
#define prime2 1e9+9
#define pi 3.14159265
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define scand(x) scanf("%llf",&x)
#define f(i,a,b) for(int i=a;i<=b;i++)
#define scan(a) scanf("%d",&a)
#define mp(a,b) make_pair((a),(b))
#define P pair<int,int>
#define dbg(args) cout<<#args<<":"<<args<<endl;
#define inf 0x3f3f3f3f
const int maxn=1e7+;
int n,m,t;
inline int read(){
int ans=,w=;
char ch=getchar();
while(!isdigit(ch)){if(ch=='-')w=-;ch=getchar();}
while(isdigit(ch))ans=(ans<<)+(ans<<)+ch-'',ch=getchar();
return ans*w;
}
int f[maxn],cnt[maxn],ans=;
void init()
{
f(i,,maxn-)f[i]=i;
f(i,,maxn-)cnt[i]=;
ans=;
}
int find(int x)
{
if(x==f[x])return x;
return f[x]=find(f[x]);
}
void Union(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx==fy)return ;
else
{
f[fx]=fy;
cnt[fy]+=cnt[fx];
}
}
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
std::ios::sync_with_stdio(false);
while(scan(m)!=EOF)
{
int a,b;
init();
if(m==)
{
pf("1\n");
continue;
}
int maxr=;
f(i,,m)
{
a=read(),b=read(),Union(a,b);
maxr=max(maxr,max(a,b));
}
f(i,,maxr)ans=max(ans,cnt[i]);
pf("%d\n",ans);
}
}