CodeForces - 767C Garland（乱搞）

最新推荐文章于 2018-11-15 21:47:15 发布

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最新推荐文章于 2018-11-15 21:47:15 发布

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分类专栏：思维题／模拟／乱搞／贪心

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这篇博客讲述了Dima在梦中得到的一个特殊圣诞花环，花环由灯泡组成，每两个灯泡间由电线直接或间接连接。Dima想要将花环割成三部分，每部分灯泡亮度（温度之和）相等。他需要找到两条电线来割断，使得这个目标能实现。博客提供了一个n个灯泡的输入示例，每个灯泡有一个挂靠的灯泡编号（0表示没有）和温度。博主通过计算每个节点的累计温度，采用后序遍历的方法寻找解决方案。如果找到的节点数大于等于3，说明存在分割方案，并给出任意一组可行的答案。

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题目链接：http://codeforces.com/problemset/problem/767/C点击打开链接

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.

$\text{[math]}$

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.

Input

The first line contains single integer n (3 ≤ n ≤ 106) — the number of lamps in the garland.

Then n lines follow. The i-th of them contain the information about the i-th lamp: the number lamp ai, it is hanging on (and 0, if is there is no such lamp), and its temperature ti ( - 100 ≤ ti ≤ 100). The lamps are numbered from 1 to n.

Output

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.

   Examples
 

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

     output
   
1 4

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

     output
   
-1

Note

The garland and cuts scheme for the first example:

$\text{[math]}$

先把每个节点的值计算总和然后后序遍历将值从下往上累加碰到等于总和/3的就将当前节点他加入容器中

如果容器中的数目大于等于三说明存在能够分成三段的多种情况

取出前面两个就行

#include <iostream>
#include <queue>
#include <stdio.h>
#include <stdlib.h>
#include <stack>
#include <limits.h>
#include <string>
#include <string.h>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
using namespace std;
vector < int > root[1111111];
int temp[1111111];
int be;
int sum=0;
vector<int > ans;
int getsum(int now)
{
    for(int i=0;i<root[now].size();i++)
    {
        getsum(root[now][i]);
    }
    for(int i=0;i<root[now].size();i++)
        temp[now]+=temp[root[now][i]];
    if(temp[now]==sum)
       {
           ans.push_back(now);
           temp[now]-=sum;
       }
}
int main()
{
    int n=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        int m,k;
        scanf("%d%d",&m,&k);
        root[m].push_back(i);
        temp[i]=k;
        if(m==0)
            be=i;
        sum+=k;
    }

    if(sum%3!=0)
        cout << "-1";
    else
    {
        sum/=3;
        getsum(be);
        //cout << be <<endl;
        if(ans.size()>=3)
            {
                for(int i=0;i<2;i++)
                        cout << ans[i] <<" ";
            }
        else
            cout << "-1";
    }
}