Apple Tree

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 22613 Accepted: 6875

Description

There is an apple tree outside of kaka’s house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.

The tree has N forks which are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won’t grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.

The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?

Input

The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree.

The following N - 1 lines each contain two integers u and v, which means fork u and fork v are connected by a branch.

The next line contains an integer M (M ≤ 100,000).

The following M lines each contain a message which is either

“C x” which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork.

or

“Q x” which means an inquiry for the number of apples in the sub-tree above the fork x, including the apple (if exists) on the fork x

Note the tree is full of apples at the beginning

Output

For every inquiry, output the correspond answer per line.

Sample Input

3

1 2

1 3

3

Q 1

C 2

Q 1

Sample Output

3

2

题意:

就是给你一棵树,每一个结点一开始都有一个苹果,然后有两个操作,一个是改变一个结点的值。即有苹果变无苹果,无苹果变有苹果。另一个是求一个结点下面的子树有多少个苹果。这个题目,询问次数100000,显然要用树状数组。属于更新点,求区间和。要知道树状数组无非两种类型,更新点求区间,更新区间求点。此外还要会dfs序列。具体见代码里dfs()函数。dfs序列可以自己百度一下是什么。这道题目是最基础的在树的结构里用树状数组统计

#include <iostream>
#include <string.h>
#include <math.h> using namespace std; #define MAX 100010
int head[MAX];
int c[MAX];
int apple[MAX];
int tol;
int n,m;
int tag;
int _right[MAX];
int _left[MAX];
int ans;
int vis[MAX];
struct Node
{
int value;
int next;
}edge[MAX*2];
void modify(int x,int y)
{
edge[tol].value=y;
edge[tol].next=head[x];
head[x]=tol++;
}
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int num)
{
while(x<=n)
{
c[x]+=num;
x+=lowbit(x);
}
}
int sum(int x)
{
int _sum=0;
while(x>0)
{
_sum+=c[x];
x-=lowbit(x);
}
return _sum;
}
void dfs(int x)
{
vis[x]=1;
tag++;
_left[x]=tag;
int a=head[x];
while(a!=-1)
{
if(vis[edge[a].value]==0) dfs(edge[a].value); a=edge[a].next;
}
_right[x]=tag;
}
void init()
{
memset(c,0,sizeof(c));
memset(head,-1,sizeof(head));
memset(vis,0,sizeof(vis));
}
int main()
{
int x,y;
char a;
int b;
scanf("%d",&n);
tol=0;
tag=0;
init();
for(int i=1;i<=n;i++)
{
update(i,1);
apple[i]=1;
}
for(int i=0;i<n-1;i++)
{
scanf("%d%d",&x,&y);
modify(x,y);
modify(y,x);
}
dfs(1);
scanf("%d",&m);
for(int i=0;i<m;i++)
{
getchar();
scanf("%c",&a);
scanf("%d",&b);
if(a=='C')
{
if(apple[b]==1)
{
update(_left[b],-1);
apple[b]=0;
}
else
{
update(_left[b],1);
apple[b]=1;
}
}
else
{
ans=sum(_right[b])-sum(_left[b]-1);
printf("%d\n",ans);
} }
return 0;
}

POJ--3321 Apple Tree(树状数组+dfs(序列))的更多相关文章

  1. POJ 3321 Apple Tree (树状数组+dfs序)

    题目链接:http://poj.org/problem?id=3321 给你n个点,n-1条边,1为根节点.给你m条操作,C操作是将x点变反(1变0,0变1),Q操作是询问x节点以及它子树的值之和.初 ...

  2. POJ 3321 Apple Tree 树状数组+DFS

    题意:一棵苹果树有n个结点,编号从1到n,根结点永远是1.该树有n-1条树枝,每条树枝连接两个结点.已知苹果只会结在树的结点处,而且每个结点最多只能结1个苹果.初始时每个结点处都有1个苹果.树的主人接 ...

  3. POJ 3321 Apple Tree(树状数组)

                                                              Apple Tree Time Limit: 2000MS   Memory Lim ...

  4. POJ 3321 Apple Tree 树状数组 第一题

    第一次做树状数组,这个东西还是蛮神奇的,通过一个简单的C数组就可以表示出整个序列的值,并且可以用logN的复杂度进行改值与求和. 这道题目我根本不知道怎么和树状数组扯上的关系,刚开始我想直接按图来遍历 ...

  5. E - Apple Tree(树状数组+DFS序)

    There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. ...

  6. 3321 Apple Tree 树状数组

    LIANJIE:http://poj.org/problem?id=3321 给你一个多叉树,每个叉和叶子节点有一颗苹果.然后给你两个操作,一个是给你C清除某节点上的苹果或者添加(此节点上有苹果则清除 ...

  7. POJ 3321:Apple Tree 树状数组

    Apple Tree Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 22131   Accepted: 6715 Descr ...

  8. POJ3321 Apple Tree(树状数组)

    先做一次dfs求得每个节点为根的子树在树状数组中编号的起始值和结束值,再树状数组做区间查询 与单点更新. #include<cstdio> #include<iostream> ...

  9. POJ 2486 Apple Tree [树状DP]

    题目:一棵树,每个结点上都有一些苹果,且相邻两个结点间的距离为1.一个人从根节点(编号为1)开始走,一共可以走k步,问最多可以吃多少苹果. 思路:这里给出数组的定义: dp[0][x][j] 为从结点 ...

随机推荐

  1. Go语言的类型转换和类型断言

    https://my.oschina.net/chai2010/blog/161418 https://studygolang.com/articles/9335  类型转换.类型断言和类型切换 ht ...

  2. js提取新浪邮箱的信用卡

    js提取用户新浪邮箱中的信用卡信息,是js非nodejs. 对比py,之前就做不好,出现了复杂点选验证码.js的开发速度只需要py的三分之一,甚至十分之一. js在客户端执行,py在后端执行,py要实 ...

  3. Oauth2.0(五):Authorization Code 授权

    Authorization Code 方式适用于有自己的服务器的应用.之所以叫这个名字,是因为流程中引入了一个叫做 authorization code 的东西.这个东西是一个一次性的临时凭证,用来换 ...

  4. IIS------如何安装IIS

    1.打开“控制面板”->“程序”->“打开或关闭Windows功能” 2.如图所示: 3.如图所示: 4.点击确定,配置完成

  5. Java 流(Stream)、文件(File)和IO -- Java ByteArrayInputStream类

    字节数组输入流在内存中创建一个字节数组缓冲区,从输入流读取的数据保存在该字节数组缓冲区中.创建字节数组输入流对象有以下几种方式. 接收字节数组作为参数创建: ByteArrayInputStream ...

  6. centos7 python3.5中引入sqlite3

    在centos系统中创建Django app,报错如下: django.core.exceptions.ImproperlyConfigured: Error loading either pysql ...

  7. java List分批处理

    java List分批处理,例如对List中的数据进行批量插入. 方法一: /** * ClassName:Test List分批处理 * @author Joe * @version * @sinc ...

  8. zabbix的启动和关闭脚本

    1. zabbix客户端的系统服务脚本 1.1 拷贝启动脚本 zabbix的源码提供了系统服务脚本,在/usr/local/src/zabbix-3.2.6/misc/init.d目录下,我的系统是C ...

  9. Git和GitHub入门基础

    -----------------------------------------//cd F:/learngit // 创建仓库git init  // 在当前目录下创建空的git仓库------- ...

  10. OpenCV——识别各省份地图轮廓

    好久没有发OpenCV的博客了,最近想到了一个识别地图轮廓的方案,就写来试试.(识别中国的28个省份地图轮廓,不考虑直辖市) 首先,我的基本思路是  用最小的矩形将地图的轮廓圈出来,可以根据长方形的长 ...