codeforces 622E E. Ants in Leaves(贪心+dfs)

题目链接：

E. Ants in Leaves

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Tree is a connected graph without cycles. A leaf of a tree is any vertex connected with exactly one other vertex.

You are given a tree with n vertices and a root in the vertex 1. There is an ant in each leaf of the tree. In one second some ants can simultaneously go to the parent vertex from the vertex they were in. No two ants can be in the same vertex simultaneously except for the root of the tree.

Find the minimal time required for all ants to be in the root of the tree. Note that at start the ants are only in the leaves of the tree.

Input

The first line contains integer n (2 ≤ n ≤ 5·105) — the number of vertices in the tree.

Each of the next n - 1 lines contains two integers xi, yi (1 ≤ xi, yi ≤ n) — the ends of the i-th edge. It is guaranteed that you are given the correct undirected tree.

Output

Print the only integer t — the minimal time required for all ants to be in the root of the tree.

Examples

input

12
1 2
1 3
1 4
2 5
2 6
3 7
3 8
3 9
8 10
8 11
8 12

output

6

input

2
2 1

output

1

题意：

除根节点外每个节点只能容纳一只蚂蚁,现在每个叶子节点都有一个蚂蚁,问这些蚂蚁都到达根节点的最短时间是多少;

思路：

贪心,深度小的蚂蚁一定要先走才能保证时间最短,所有按深度排序后每个叶子节点的蚂蚁到达根节点的时间就是上一个节点的最大时间+1与其深度的相比的较大值,即ans[i]=max(ans[i-1]+1,dep[i]);

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=5e5+10;
const int maxn=(1<<20)+14;
const double eps=1e-12;

int n,cnt;
vector<int>ve[N];
int dep[N],num[N];
void dfs(int cur,int fa,int deep)
{
    int len=ve[cur].size();
    for(int i=0;i<len;i++)
    {
        int x=ve[cur][i];
        if(x==fa)continue;
        dfs(x,cur,deep+1);
    }
    if(len==1)dep[++cnt]=deep;
}
int main()
{
    int u,v;
    read(n);
    For(i,1,n-1)
    {
        read(u);read(v);
        ve[u].push_back(v);
        ve[v].push_back(u);
    }
    int len=ve[1].size(),ans=0;
    for(int i=0;i<len;i++)
        {
            cnt=0;
            dfs(ve[1][i],1,1);
            sort(dep+1,dep+cnt+1);
            for(int j=1;j<=cnt;j++)
            {
                num[j]=max(num[j-1]+1,dep[j]);
            }
            ans=max(ans,num[cnt]);
        }
    //for(int i=0;i<len;i++)ans=max(ans,);
    print(ans);
    return 0;
}