CF1042F Leaf Sets

传送门

Luogu

Solution

考虑将每一个子树分开处理，那么对于子树内距离(ge K)的就分一块，然后这样子贪心不断往上走就是对的我也不知道为什么

大致可以类比一下树分块的做法。

Code

/*
  mail: mleautomaton@foxmail.com
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define REP(a,b,c) for(int a=b;a<=c;a++)
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=1000010;
int n,k,d[N],ans;
vector<int>G[N];
int dfs(int u,int ff){
	if(d[u]==1)return 0;
	int top=0;vector<int>sta;
	for(int v:G[u]){if(v==ff)continue;sta.push_back(dfs(v,u)+1);}
	sort(sta.begin(),sta.end());
	int now=sta.size()-1;
	for(;now;now--){
		if(sta[now]+sta[now-1]<=k)break;
		ans++;
	}
	return sta[now];
}
int main(){
	n=gi();k=gi();
	for(int i=1;i<n;i++){int u=gi(),v=gi();G[u].push_back(v);G[v].push_back(u);d[u]++;d[v]++;}
	int rt=1;
	for(int i=2;i<=n;i++)if(d[i]>1){rt=i;break;}
	dfs(rt,rt);
	printf("%d
",ans+1);
	return 0;
}