ABC-385-E - Snowflake Tree · znzryb

## 思路讲解

总体上来说这题还是比较顺的，就是顺着做就好了，也没用什么算法，for循环就完了

```cpp
		sort(all1(sons),greater<ll>());
		FOR(j, 0, sons.size()-1){
			remainNode=max(remainNode,1+sons[j]*(j+1)+j+1);
		}
```

核心是这一段代码，因为是0-based，所以要 j+1

## AC代码

```cpp
#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x,size) x.begin()+1,x.begin()+size+1
#define all1(x) x.begin(),x.end()

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(3e5)+10,MAXval=static_cast<ll>(1e18)+3;

ll N,T;
vector<ll> G[MAXN];

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>N;
	FOR(i, 1, N-1){
		ll u,v;
		cin>>u>>v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	ll ans=MAXval;
	FOR(i, 1, N){
		ll remainNode=0;
		vector<ll> sons;	// 他的子节点的子节点有几个？
		FOR(j, 0, G[i].size()-1){
			ll node=G[i][j];
			sons.push_back(G[node].size()-1);
		}
		sort(all1(sons),greater<ll>());
		FOR(j, 0, sons.size()-1){
			remainNode=max(remainNode,1+sons[j]*(j+1)+j+1);
		}
		ans=min(ans,N-remainNode);
		
//		cout<<"debug:\n";
//		FOR(j, 0, sons.size()-1){
//			cout<<sons[j]<<' ';
//		}
//		cout<<'\n';
	}
	cout<<ans<<'\n';
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc385/submissions/62235314
*/
```

## 心路历程（WA，TLE，MLE……）