链接：​​https://www.nowcoder.com/acm/contest/157/B​​

本来是想要模拟节点的收缩过程，但是实在是太难做了。后来看了别人的代码才想到，其实跟节点的分布没有关系，只跟与一节点相连的点数有关系（设为X），一秒最多有X个人到达1。可以用并查集，但是我头铁就是要dfs，vector超时了，用邻接表结构体少开了一倍又re了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
#define mem(a,x) memset(a,x,sizeof(a))
#define s1(x) scanf("%d",&x)
#define s2(x,y) scanf("%d%d",&x,&y)
#define s3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s4(x,y,z,k) scanf("%d%d%d%d",&x,&y,&z,&k)
#define ff(a,n) for(int i = 0 ; i < n; i++) scanf("%d",a+i)
#define tp(x) printf("x = %d\n",x)
#define ansp(x) printf("%d\n",x)
//inline ll ask(int x){ll res=0;while(x)res+=c[x],x-=x&(-x);return res;}
//inline void add(int x,int d){while(x<=n)c[x]+=d,x+=x&(-x);}
#define ls 2*rt
#define rs 2*rt+1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define ll long long
using namespace std;
typedef pair<int,int> pii;
const ll inf = 0x3f3f3f3f;
 
const int mx = 1e6+10;
//vector<int>ve[mx];
 
bool vis[mx];
int te,ans,n,head[mx],tol;
 
struct{
    int to,ne;
}p[mx*2];
 
inline void ad(int a, int b){
    //tol++;
    //p[tol].from = a;
    p[tol].to = b;
    p[tol].ne = head[a];
    head[a] = tol++;   
}
void dfs(int x){
    vis[x] = 1;
    te++;
    int v;
    for(int i = head[x]; i != -1 ; i = p[i].ne){
        v = p[i].to;
        if(!vis[v])
            dfs(v);
    }
}
int main(){
    //  freopen("F:\\in.txt","r",stdin);
    //int T=10; scanf("%d",&T);
    //s1(n);
    scanf("%d",&n);
    tol = 1;
    mem(head,-1);
    int a,b;
    for(int i=1; i < n; i++){
        //s2(a,b);
        scanf("%d%d",&a,&b);
    //  if(a > 1e6+10|| b>1e6+10)
    //      while(1);
        ad(a,b);
        ad(b,a);
    }
    ans = 0;
    vis[1] = 1;
    //int v;
    for(int i = head[1]; i != -1 ; i = p[i].ne){
        //v = p[i]
        te = 0;
        dfs(p[i].to);
        ans = max(ans,te);
    }
    printf("%d\n",ans);
 
 
    return 0;
}