POJ 3321 DFS序+线段树

单点修改树中某个节点,查询子树的性质.DFS序 子树序列一定在父节点的DFS序列之内,所以可以用线段树维护.

/*
DFS序 +线段树
*/

#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
#include <iostream>
using namespace std;

#define LL(a)   a<<1
#define RR(a)   a<<1|1

const int MaxL = 200002;

int pre[MaxL];  // first travel
int nxt[MaxL];  // nxt travel
bool vis[MaxL];
int N;
vector<vector<int> > E(MaxL);

struct Seg_tree
{
    int left, right;
    int sum;
} tt[MaxL<<2];


void PushUp(int idx)
{
    tt[idx].sum = tt[LL(idx)].sum + tt[RR(idx)].sum;
}

void build(int l,int r,int idx)
{
    tt[idx].left = l, tt[idx].right = r;
    tt[idx].sum = r-l+1;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(l,m, LL(idx));
    build(m+1, r, RR(idx));
}

void update(int p, int idx = 1)
{
    if(p == tt[idx].left && p == tt[idx].right)
    {
        tt[idx].sum ^=1;
        return ;
    }
    int mid = (tt[idx].left + tt[idx].right)>>1;
    if(p <= mid) update(p, LL(idx));
    else update(p, RR(idx));
    PushUp(idx);
}

int query(int l, int r, int idx = 1)
{
    if(l == tt[idx].left && r ==tt[idx].right)
        return tt[idx].sum;
    int mid = (tt[idx].left + tt[idx].right)>>1;
    if(r <= mid) return query(l,r, LL(idx));
    else if(l> mid) return query(l,r, RR(idx));
    else
        return query(l, mid, LL(idx))+ query(mid+1, r, RR(idx));
}

int mark = 1;
void dfs( int a)
{
    vis[a] = 1;
    pre[a] = mark++;
    for(int i=0; i<E[a].size(); i++)
    {
        if(!vis[E[a][i]])
            dfs(E[a][i]);
    }
    nxt[a] = mark++;
}
int main()
{
//    freopen("1.txt","r",stdin);
    memset(pre, 0, sizeof(pre));
    memset(nxt, 0 ,sizeof(nxt));
    memset(vis, 0 ,sizeof(vis));

    scanf("%d", &N);
    for(int i=1; i<N; i++)
    {
        int a,b;
        scanf("%d%d", &a,&b);
        E[a].push_back(b);
        E[b].push_back(a);
    }
    dfs(1);

    N*=2;
    build(1, N, 1);
    int M;
    scanf("%d", &M);
    for(int i=1; i<=M; i++)
    {
        char c;
        int a;
        scanf("%s", &c);
        scanf("%d",&a);
        if(c =='Q')
        {
            cout<<query(pre[a],nxt[a], 1) /2<<endl;
        }
        else
        {
            update(pre[a],1);
            update(nxt[a],1);
        }
    }
    return 0;
}