CF767C Garland (#图论)

题意翻译

nn个节点的树

第ii个节点权值为a_iai​. n<=10^6n<=106

-100<=a_i<=100−100<=ai​<=100

问是否能够删除掉两条边,使得该树分成三个不为空,并且每部分权值之和相等.

无解输出-1−1 否则输出要删除边(u->vu−>v)的vv节点序号.

题目描述

Once at New Year Dima had a dream in which he was presented a fairy garland. A garland is a set of lamps, some pairs of which are connected by wires. Dima remembered that each two lamps in the garland were connected directly or indirectly via some wires. Furthermore, the number of wires was exactly one less than the number of lamps.

There was something unusual about the garland. Each lamp had its own brightness which depended on the temperature of the lamp. Temperatures could be positive, negative or zero. Dima has two friends, so he decided to share the garland with them. He wants to cut two different wires so that the garland breaks up into three parts. Each part of the garland should shine equally, i. e. the sums of lamps' temperatures should be equal in each of the parts. Of course, each of the parts should be non-empty, i. e. each part should contain at least one lamp.

Help Dima to find a suitable way to cut the garland, or determine that this is impossible.

While examining the garland, Dima lifted it up holding by one of the lamps. Thus, each of the lamps, except the one he is holding by, is now hanging on some wire. So, you should print two lamp ids as the answer which denote that Dima should cut the wires these lamps are hanging on. Of course, the lamp Dima is holding the garland by can't be included in the answer.

输入格式

The first line contains single integer nn ( 3<=n<=10^{6}3<=n<=106 ) — the number of lamps in the garland.

Then nn lines follow. The ii -th of them contain the information about the ii -th lamp: the number lamp a_{i}ai​ , it is hanging on (and 00 , if is there is no such lamp), and its temperature t_{i}ti​ ( -100<=t_{i}<=100−100<=ti​<=100 ). The lamps are numbered from 11 to nn .

输出格式

If there is no solution, print -1.

Otherwise print two integers — the indexes of the lamps which mean Dima should cut the wires they are hanging on. If there are multiple answers, print any of them.

输入输出样例

输入 #1复制

6
2 4
0 5
4 2
2 1
1 1
4 2

输出 #1复制

1 4

输入 #2复制

6
2 4
0 6
4 2
2 1
1 1
4 2

输出 #2复制

-1

说明/提示

The garland and cuts scheme for the first example:

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思路

为什么题目标签写的是树形dp？这不就是个图论吗。。要硬说是树形dp的话也勉强像吧。。

题意：将一颗树的边分成三棵子树，使每一颗子树的权值和大小相等，输出要断开的边的编号，如果不能构造出合法的子树，就输出-1。如果有多个答案输出任意一对。

#include 
#include 
#define maxn 1000001
using namespace std;
int n,m,s[maxn],cnt,head[maxn],sum,dp[maxn],value[maxn],root,pos;
struct node
{
	int nxt,to;
}e[maxn<<1];
inline void add(int u,int v)
{
	e[++cnt].to=v;
	e[cnt].nxt=head[u];
	head[u]=cnt;
}
void dfs(int i,int fa)
{
	register int j,v;
	dp[i]=value[i];
	for(j=head[i];j;j=e[j].nxt)
	{
		int v(e[j].to);//子节点 
		if(v==fa) continue;
		dfs(v,i);
		dp[i]+=dp[v];//记录权值和 
	}
	if(dp[i]==sum)//如果等于三分之一sum 
	{
		s[++pos]=i;
		dp[i]=0;
		return;
	}
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	register int i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	{
		int u;
		cin>>u>>value[i];
		if(u==0)
		{
			root=i;
		}
		else
		{
			add(i,u);
			add(u,i);
		}
		sum+=value[i];
	}
	if(sum%3)
	{
		cout<<-1<