POJ3321Apple Tree[樹轉序列 BIT]

Candy?發表於2016-09-19
APP
Apple Tree
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 26995   Accepted: 8007

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

Source

POJ Monthly--2007.08.05, Huang, Jinsong
 
題意：加減一個節點的蘋果(異或) 查詢以x為根的子樹的蘋果數
求dfs序(先序遍歷)，每個子樹就是序列上的一段，單點修改區間查詢
//
//  main.cpp
//  poj3211
//
//  Created by Candy on 9/19/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+5,M=1e5+5,INF=1e9;
int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}
int n,m,x;
char c[5];
struct edge{
    int v,ne;
}e[N<<1];
int cnt=0,h[N];
inline void ins(int u,int v){
    cnt++;
    e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;
}
int a[N],p=0,st[N],ed[N];
void dfs(int u,int fa){
    p++;st[u]=p;
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v;
        if(v==fa) continue;
        dfs(v,u);
    }
    ed[u]=p;
}
int bit[N];
inline int lowbit(int x){
    return x&-x;
}
void buildBit(){
    for(int i=1;i<=n;i++){
        a[i]=1;
        bit[i]+=a[i];
        if(i+lowbit(i)<=n) bit[i+lowbit(i)]+=bit[i];
    }
}
inline void add(int x,int d){
    while(x<=n){
        bit[x]+=d;x+=lowbit(x);
    }
}
inline int sum(int x){
    int ans=0;
    while(x>0){
        ans+=bit[x];x-=lowbit(x);
    }
    return ans;
}
inline int query(int l,int r){
    return sum(r)-sum(l-1);
}
int main(int argc, const char * argv[]) {
    n=read();
    for(int i=1;i<=n-1;i++) ins(read(),read());
    dfs(1,0);
    buildBit();
    m=read();
    //for(int i=1;i<=n;i++) printf("%d %d  %d %d\n",st[i],ed[i],a[i],bit[i]);
    for(int i=1;i<=m;i++){
        scanf("%s",c);
        if(c[0]=='C'){
            x=read();int b=st[x];
            if(a[b]) add(b,-1);
            else add(b,1);
            a[b]^=1;
        }
        if(c[0]=='Q'){
            x=read();
            printf("%d\n",query(st[x],ed[x]));
        }
    }
    return 0;
}

 

 

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