D. Peculiar apple-tree
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n. Inflorescence number 1 is situated near base of tree and any other inflorescence with number i (i > 1) is situated at the top of branch, which bottom is pi-th inflorescence and pi < i.
Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a-th inflorescence gets to pa-th inflorescence. Apples that end up in first inflorescence are gathered by Arcady in exactly the same moment. Second peculiarity of this tree is that once two apples are in same inflorescence they annihilate. This happens with each pair of apples, e.g. if there are 5 apples in same inflorescence in same time, only one will not be annihilated and if there are 8 apples, all apples will be annihilated. Thus, there can be no more than one apple in each inflorescence in each moment of time.
Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.
Input
First line of input contains single integer number n (2 ≤ n ≤ 100 000) — number of inflorescences.
Second line of input contains sequence of n - 1 integer numbers p2, p3, ..., pn (1 ≤ pi < i), where pi is number of inflorescence into which the apple from i-th inflorescence rolls down.
Output
Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.
Examples
input
Copy
3
1 1
output
1
input
Copy
5
1 2 2 2
output
3
input
Copy
18
1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4
output
4
Note
In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won't be able to collect them.
In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.
题意：
一棵苹果树，有n个花蕊。花蕊成树状分布。一天果子全部成熟了，于是开始下落，每秒下落一个节点。神奇之处在于同一个节点同一时刻的果子会两两相消，即只能存k%2个果子。求可以落到地面多少果子。输入每个花蕊的前驱。
思路：入眼就是感觉应该用树进行操作，但思考后发现除了长得是树形，别的没啥关系，原本是使用暴力推解，但是发现这是动态的过程，所以就联想到dp，但是在思路还没缕清的时候，比赛截止了，之后又去百度题解，这个题目既可以dp也可以dfs
先讲一下dp做法：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<stack>
using namespace std;
int nextt[100005];
int level[100005];
int dp[100005];
int anser = 0;
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(dp,0,sizeof(dp));
memset(level,0,sizeof(level));
anser=0;
level[1] = dp[1] = 1;
for (int i = 2; i <= n; i++)
{
scanf("%d", &nextt[i]);//存放它下一层节点的位置 ，存放关系
dp[i] = dp[nextt[i]] + 1;//dp用来存放由下层向上层递推得到这层的果子的数目
level[dp[i]]++;//累计这个点实际能有多少到达底部的数目
}
for(int i=1;i<=100000;i++)
{
if(level[i]&1)//判断奇偶 ，奇数加一偶数归零
anser++;
}
// for(int i=1;i<=n;i++)
// printf("%d %d %d\n",nextt[i],dp[i],level[i]);
cout<<anser<<endl;
}
return 0;
}很巧妙的利用了递推关系，然后level中存放最后能到底部的数量，最后取余求和就得到最后结果。