POJ 3321 Apple Tree (树状数组)

题意：有n个苹果被树枝连接，这是一棵树！有两种操作C ind 和 Q ind，前者是摘下ind苹果，如果没有，哪么会长出新的一个，后者是查询ind有几个子苹果。

思路：第二道树状数组，自己想了很久不知道怎么转化，原来是利用树的性质，dfs一遍，记录每个节点的low和high值，那么他的子结点的low值和high值肯定在[low,high]之间，然后就可以通过tre[high[i]]-tre[low[i]-1]来查询当前节点的子结点个数了，汗一个，又不是自己的......

/* Apple tree
 * 树状数组第二题，转化很经典，将树转化为数组，low[],high[]
 * 记录节点位置，树状数组求和修改O(log(n))
 */
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
const int MAXSIZE = 37;
const int MAXN = 202000;
const int INF = 1000000000;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

typedef struct{
    int v,next;
}Edge;

Edge edge[MAXN*20];
int tre[MAXN],low[MAXN],net[MAXN],high[MAXN];
int n_tre,n,cnt,index,m;
bool visit[MAXN];

void add_edge(const int& u,const int& v)
{
    edge[index].v = v;
    edge[index].next = net[u];
    net[u] = index;
    ++index;
    edge[index].v = u;
    edge[index].next = net[v];
    net[v] = index;
    ++index;
}
int init()
{
    cnt = 0;
    index = 0;
    CLR(tre,0);
    CLR(visit,0);
    CLR(net,-1);
    return 0;
}
int lowbit(int x)
{
    return x&(-x);
}
void modify(int ind,int delta)
{
    while( ind <= n_tre)
    {
        tre[ind] += delta;
        ind += lowbit(ind);
    }
}
int get_sum(int ind)
{
    int sum = 0;
    while(ind > 0)
    {
        sum += tre[ind];
        ind -= lowbit(ind);
    }
    return sum;
}
int input()
{
    int i,a,b;
    for(i = 1; i < n; ++i)
    {
        scanf("%d %d",&a,&b);
        add_edge(a,b);
    }
    return 0;
}
void dfs(const int& u)
{
    ++cnt;
    visit[u] = true;
    low[u] = cnt;
    for(int i = net[u]; i != -1; i = edge[i].next)
        if(!visit[edge[i].v])
            dfs(edge[i].v);
    high[u] = cnt;
}
int work()
{
    int i,j,tmp,ind;
    int ans;
    char c[10];
    n_tre = n<<1;
    dfs(1);
    IN(m);
    CLR(visit,0);
    for(i = 0; i < m; ++i)
    {
        scanf("%s",c);
        scanf("%d",&ind);
        if(c[0] == 'Q')
        {
            ans =high[ind] - low[ind] + 1 + 
                get_sum(high[ind]) - get_sum(low[ind]-1);
            OUT(ans);
        }else
        {
            if(visit[ind])
                modify(low[ind],1);
            else
                modify(low[ind],-1);
            visit[ind] ^= 1;
        }
    }
    return 0;
}
int main()
{
    IN(n);
    {
        init();
        input();
        work();
    }
    return 0;
}