题目简单,不多解释。

Code:

#include<cstdio>
#include<queue>
using namespace std;
const int maxn = 1000000 + 3;
int head[maxn], to[maxn], nex[maxn], cnt, dep[maxn], numv[maxn];
queue<int>Q;
inline void add_edge(int u,int v)
{
nex[++cnt] = head[u], head[u] = cnt, to[cnt] = v;
}
int main()
{
//freopen("input.in","r",stdin);
int n;
scanf("%d",&n);
for(int i = 2;i <= n; ++i)
{
int a; scanf("%d",&a);
add_edge(a,i);
}
Q.push(1); dep[1] = 1; numv[1] = 1;
while(!Q.empty())
{
int u = Q.front(); Q.pop();
for(int v = head[u]; v ; v = nex[v])
{
dep[to[v]] = dep[u] + 1;
++numv[dep[to[v]]];
Q.push(to[v]);
}
}
int ans = 0;
for(int i = 1;i <= 1000000; ++i)
{
if(numv[i] % 2 == 1) ++ans;
}
printf("%d",ans);
return 0;
}

Codeforces Round #468 (Div. 2 )D. Peculiar apple-tree_BFS的更多相关文章

  1. Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)D. Peculiar apple-tree

    In Arcady's garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity ...

  2. Codeforces Round #468 Div. 2题解

    A. Friends Meeting time limit per test 1 second memory limit per test 256 megabytes input standard i ...

  3. Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)B. World Cup

    The last stage of Football World Cup is played using the play-off system. There are n teams left in ...

  4. Codeforces Round #468 Div. 1

    D:首先考虑如果给定白棋位置,如何判断胜负.黑棋获胜需要四个方向都有能贴上白棋的棋子.由于每一轮都必须移动,显然先对平面黑白染色一下,只有与白棋所在格异色的黑棋才需要考虑.考虑让一个黑棋去贴上白棋某个 ...

  5. Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)

    A.B都是暴力搞一搞. A: #include<bits/stdc++.h> #define fi first #define se second #define mk make_pair ...

  6. codeforces 930b//Game with String// Codeforces Round #468 (Div. 1)

    题意:一个串,右循环移位后,告诉你第一个字母,还能告诉你一个,问你能确定移位后的串的概率. 用map记录每个字母出现的位置.对于每个字母,用arr[j][k]记录它的所有出现位置的后j位是字母k的个数 ...

  7. Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)C. Laboratory Work

    Anya and Kirill are doing a physics laboratory work. In one of the tasks they have to measure some v ...

  8. Codeforces Round #468 (Div. 2, based on Technocup 2018 Final Round)A. Friends Meeting

    Two friends are on the coordinate axis Ox in points with integer coordinates. One of them is in the ...

  9. Codeforces Round #257 (Div. 1)A~C(DIV.2-C~E)题解

    今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and ...

随机推荐

  1. Java常用工具类---XML工具类、数据验证工具类

    package com.jarvis.base.util; import java.io.File;import java.io.FileWriter;import java.io.IOExcepti ...

  2. linux 性能分析与优化

    一.影响Linux服务器性能的因素 1.操作系统级 (CPU 内存 磁盘I/O性能 网络带宽) 2.程序应用级 二.系统性能评估标准   好  坏 极差 cpu user% +sys% <70% ...

  3. python与图灵机器人交互(ITCHAT版本)

    #!/usr/bin/env python#-*- coding:utf-8 -*- @Author : wujf @Time:2018/9/5 17:42import requestsimport ...

  4. Linux进程地址管理之mm_struct

    FROM : http://www.cnblogs.com/Rofael/archive/2013/04/13/3019153.html Linux对于内存的管理涉及到非常多的方面,这篇文章首先从对进 ...

  5. Statement对象sql注入漏洞的问题

    现在通过mysql以及oracle来测试sql注入  漏洞 mysql中的注释#    oracle中的注释为-- 所以注入漏洞就产生了 //登录测试 public void login()throw ...

  6. CodeForcesGym 100548G The Problem to Slow Down You

    The Problem to Slow Down You Time Limit: 20000ms Memory Limit: 524288KB This problem will be judged ...

  7. BA-WG-泰豪发电机

    泰豪发电机的控制主板有2个端口,一个是RS232端口,一个是RS485端口,通常接网关需要将这个RS485的端口调整为modbus协议输出,再将modbus协议通过网关转换为bacnet / ip协议 ...

  8. CF49A Sleuth

    CF49A Sleuth 题目描述 Vasya plays the sleuth with his friends. The rules of the game are as follows: tho ...

  9. BAT常问问题总结以及回答(问题汇总篇)

    几个大厂的面试题目目录: java基础(40题)https://www.cnblogs.com/television/p/9397968.html 多线程(51题) 设计模式(8点) JVM(12题) ...

  10. 循环神经网络(RNN, Recurrent Neural Networks)介绍

    原文地址: http://blog.csdn.net/heyongluoyao8/article/details/48636251# 循环神经网络(RNN, Recurrent Neural Netw ...