CF1042F Leaf Sets

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原创最新推荐文章于 2026-07-31 15:57:43 发布 415 阅读 AI 写同款· GEO 优化 ›

【信息科学与工程学】计算机科学与技术——第七十七篇系统架构设计02 编号类型模型配方企业资本运作的方法/规则/条件/依据及所有的步骤（风险处理/异常处理/危机应对关联知识法律法规和裁决依据由发起人（通常是知名投资人、投行或产业资本）先设立一个只有现金、无实际业务的特殊目的收购公司（SPAC），通过IPO募集资金并在交易所上市。然后在规定时间内（通常24个月）寻找一家非上市公司进行合并（De-SPAC），使目标公司迅速实现上市，规避传统IPO的复杂流程和不确定性。1.：在开曼或特拉华注册空壳公司，注入少量种子资金。2. 阅读详情

原题链接：http://codeforces.com/contest/1042/problem/F

Leaf Sets

You are given an undirected tree, consisting of $n$ vertices.

The vertex is called a leaf if it has exactly one vertex adjacent to it.

The distance between some pair of vertices is the number of edges in the shortest path between them.

Let’s call some set of leaves beautiful if the maximum distance between any pair of leaves in it is less or equal to $k$ .

You want to split all leaves into non-intersecting beautiful sets. What is the minimal number of sets in such a split?

Input

The first line contains two integers $n$ and $k (3≤n≤10^6, 1≤k≤10^6)$ — the number of vertices in the tree and the maximum distance between any pair of leaves in each beautiful set.

Each of the next $n - 1$ lines contains two integers $v_i$ and $u_i (1≤v_i,u_i≤n)$ — the description of the $i$ -th edge.

It is guaranteed that the given edges form a tree.

Output

Print a single integer — the minimal number of beautiful sets the split can have.

Examples

input

9 3
1 2
1 3
2 4
2 5
3 6
6 7
6 8
3 9

output

input

5 3
1 2
2 3
3 4
4 5

output

input

6 1
1 2
1 3
1 4
1 5
1 6

output

Note

Here is the graph for the first example:

题解

对于每个点维护最深的叶子离自己的距离，向上合并时，将所有儿子按最深的叶子离自己的距离排序，从深到浅遍历尝试合并，如果相邻的两个距离加起来大于 $k$ ，便无法合并，向上传也没有意义，于是就将该儿子及其子树单独分成一个集合，答案+1，最后返回在删去所有分离的儿子后最深的叶节点离自己的距离。

代码

#include<bits/stdc++.h>
#define add(a,b) mmp[a].push_back(b)
using namespace std;
const int M=1e6+5;
int ans,n,k;
vector<int>mmp[M];
int dfs(int f,int v)
{
	if(mmp[v].size()==1)return 0;
	vector<int>tmp;
	for(int i=mmp[v].size()-1;i>=0;--i)if(mmp[v][i]!=f)tmp.push_back(dfs(v,mmp[v][i])+1);
	sort(tmp.begin(),tmp.end());
	for(int siz;(siz=tmp.size())>1&&tmp[siz-1]+tmp[siz-2]>k;++ans,tmp.pop_back());
	return tmp.back();
}
void in(){scanf("%d%d",&n,&k);for(int i=1,a,b;i<n;++i)scanf("%d%d",&a,&b),add(a,b),add(b,a);}
void ac(){for(int i=1;i<=n;++i)if(mmp[i].size()>1)dfs(0,i),printf("%d",ans+1),exit(0);}
int main(){in(),ac();}