題目
題解

Solution

把子樹拆成幾條鏈，每次合併短的幾條鏈

Code

#include<bits/stdc++.h>
using namespace std;
const int N=1000001;
struct node{
	int to,ne;
}e[N<<1];
int n,m,i,x,y,d[N],tot,h[N],ans;
inline char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin
 
),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int x=0,fl=1;char ch=gc();
	for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
	for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
	return x*fl;
}
void add(int x,int y){
	e[++tot]=(node){y,h[x]};
	h[x]=tot;
}
int dfs(int u,int fa){
 

	if (d[u]==1) return 0;
	vector<int>vec;
	for (int i=h[u],v;i;i=e[i].ne)
		if ((v=e[i].to)!=fa) vec.push_back(dfs(v,u)+1);
	sort(vec.begin(),vec.end());
	for (int i=vec.size()-1;i>=0;i--){
		int t=vec[i];
		if (i) t+=vec[i-1];
		if (t<=m) return vec[i];
		ans++;
	}
	return -1e9;
}
int main
 
(){
	n=rd();m=rd();
	for (i=1;i<n;i++) x=rd(),y=rd(),add(x,y),add(y,x),d[x]++,d[y]++;
	for (i=1;i<=n;i++)
		if (d[i]>1) break;
	if (dfs(i,0)>0) ans++;
	printf("%d",ans);
}