Codeforces - Garland

题目链接：Codeforces - Garland

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每个联通块的和肯定为sum/3

然后只要找到一个子树和为sum/3，那么切开。最后判断能不能切两次即可。

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AC代码：

#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N=1e6+10;
int n,a[N],sum,rt,s[N];
vector<int> g[N],v;
void dfs(int x){
	s[x]=a[x];
	for(auto to:g[x]){
		dfs(to);	s[x]+=s[to];
	}
	if(x!=rt&&s[x]*3==sum){
		v.push_back(x);	s[x]=0;
	}
}
signed main(){
	cin>>n;
	for(int i=1,fa;i<=n;i++){
		scanf("%d %d",&fa,&a[i]);	sum+=a[i];
		if(!fa)	rt=i;
		else	g[fa].push_back(i);
	}
	if(sum%3)	return puts("-1"),0;
	dfs(rt);
	if(v.size()<2)	return puts("-1"),0;
	cout<<v[0]<<' '<<v[1];
	return 0;
}