POJ 3321 Apple Tree

Apple Tree

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19316		Accepted: 5874

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

解題思路：DFS序+樹狀數組，簡而言之就是將樹狀結構利用DFS編號成序列構成一個線性結構，利用樹狀數組來維護的過程

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <bitset>
#include <cmath>
#include <utility>
#define Maxn 100005
#define Maxm 1000005
#define lowbit(x) x&(-x)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PI acos(-1.0)
#define make_pair MP
#define LL long long 
#define Inf (1LL<<62)
#define inf 0x3f3f3f3f
#define re freopen("in.txt","r",stdin)
#define wr freopen("out.txt","w",stdout)
using namespace std;
int head[Maxn],tree[Maxn],id[Maxn],st[Maxn],ed[Maxn],k,cnt;
struct
{
    int e;
    int next;
}edge[Maxn<<1];
void update(int pos,int n,int val)
{
    int i;
    for(i=pos;i<=n;i+=lowbit(i))
        tree[i]+=val;
}
int sum(int pos)
{
    int i,sum=0;
    for(i=pos;i>0;i-=lowbit(i))
        sum+=tree[i];
    return sum;
}
void add(int s,int e)
{
    edge[k].e=e;
    edge[k].next=head[s];
    head[s]=k++;
    edge[k].e=s;
    edge[k].next=head[e];
    head[e]=k++;
}
void dfs(int s,int pre)
{
    id[s]=++cnt;
    st[s]=cnt;
    for(int i=head[s];i!=-1;i=edge[i].next)
    {
        int e=edge[i].e;
        if(e==pre)
            continue;
        dfs(e,s);
    }
    ed[s]=cnt;
}
int main()
{
    int i,n,m,a,b,apple[Maxn],q,s;
    char op[3];
    //re;wr;
    while(~scanf("%d",&n))
    {
        k=cnt=0;
        memset(head,-1,sizeof(head));
        memset(tree,0,sizeof(tree));
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        dfs(1,0);
        for(i=1;i<=n;i++)
        {
            apple[i]=1;
            update(id[i],n,1);
        }
        scanf("%d",&q);
        while(q--)
        {
            scanf("%s%d",&op,&s);
            if(op[0]=='C')
            {
                apple[s]^=1;
                update(id[s],n,apple[s]?1:-1);
            }
            else
                printf("%d\n",sum(ed[s])-sum(st[s]-1));
        }
    }
    return 0;
}