CF767 C.Garland DFS

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题意：给定一棵树，每个节点拥有权值，问能否找到两个点，断开它们与父节点的边能使树分成权值和相等的三部分。权值可以为负

思路：进行两遍DFS,第一遍找最深的子树和为sum/3的节点,标记掉找到的点，同时更新剩下点的子树和，第二遍在第一遍的基础上再找一遍就可以了。注意权值可以为负，那么就意味着，虽然找到的两个点为根的子树和都等于sum/3，但不一定使根节点的子树和为sum/3，再判断一次就可以了

/** @Date    : 2017-04-18 21:27:36
  * @FileName: 767C DFS.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoundEarlf@gmail.com)
  * @Link    : https://github.com/Lweleth
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e6+20;
const double eps = 1e-8;
vectoredg[N];
int val[N];
int a[N];
bool vis[N];
int ans[2];
int sum;
int flag;
int cnt;
int rt;
int dfs(int x)
{
	int res = a[x];
	for(int i = 0; i < edg[x].size(); i++)
	{
		if(!vis[edg[x][i]])
			res += dfs(edg[x][i]);
	}
	//cout << x << "-" << res << "~"  << endl;
	if(res == sum / 3 && !flag && x != rt)
	{
		vis[x] = 1;
		flag = 1;
		//cout << x <<"@";
		ans[cnt++] = x;
		res = 0;
	}
	return val[x] = res;
}
int main()
{
	int n;
	while(~scanf("%d", &n))
	{
		int x, y;
		rt = -1;
		ans[0] = ans[1] = -1;
		sum = 0;
		for(int i = 0; i <= n; i++)
			edg[i].clear(), vis[i] = 0;

		for(int i = 1; i <= n; i++)
		{
			scanf("%d%d", &x, &y);
			a[i] = y;
			edg[x].PB(i);
			sum += y;
			if(x == 0)
				rt = i;
		}
		if(sum % 3 == 1)
		{
			printf("-1
");
			continue;
		}
		cnt = 0;
		flag = 0;
		dfs(rt);
		flag = 0;
		dfs(rt);
		if(cnt < 2 || val[rt] != sum / 3)//还要加判根节点的值 因为有可能出现负值节点
			printf("-1
");
		else 
			printf("%d %d
", ans[0], ans[1]);

	}
    return 0;
}