Water Tree 【CodeForces - 343D】【树链剖分】

题目链接

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  有N个点的树，我们有Q次操作，分别是

1. 将以v为根结点的子树全部放满为1；
2. 将v到根结点（1）的链上清空为0；
3. 查询v结点是空还是有的。

  所以，虽然挂在了Dsu on Tree的板块上晃了我一手，但还是想到了，应该考虑时间线，所以就用树链剖分来维护一下就可以了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 5e5 + 7;
int N, Q, head[maxN], cnt;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
int siz[maxN], id[maxN], tot, Wson[maxN], deep[maxN], fa[maxN], top[maxN];
void dfs_1(int u, int father)
{
    fa[u] = father; deep[u] = deep[father] + 1; siz[u] = 1;
    int maxx = 0;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == father) continue;
        dfs_1(v, u);
        siz[u] += siz[v];
        if(siz[v] > maxx)
        {
            maxx = siz[v];
            Wson[u] = v;
        }
    }
}
void dfs_2(int u, int topy)
{
    top[u] = topy;
    id[u] = ++tot;
    if(Wson[u]) dfs_2(Wson[u], topy);
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(v == fa[u] || v == Wson[u]) continue;
        dfs_2(v, v);
    }
}
int tree[maxN << 2] = {-1, -1};
inline void pushdown(int rt) { if(tree[rt]) { tree[lsn] = tree[rsn] = tree[rt]; tree[rt] = 0; } }
inline void pushup(int rt) { if(tree[lsn] && tree[lsn] == tree[rsn]) tree[rt] = tree[lsn]; }
void update(int rt, int l, int r, int ql, int qr, int val)
{
    if(ql <= l && qr >= r) { tree[rt] = val; return; }
    pushdown(rt);
    int mid = HalF;
    if(qr <= mid) update(QL, val);
    else if(ql > mid) update(QR, val);
    else { update(QL, val); update(QR, val); }
    pushup(rt);
}
int query(int rt, int l, int r, int qx)
{
    if(tree[rt] || (l == r)) return tree[rt];
    int mid = HalF;
    if(qx <= mid) return query(Lson, qx);
    else return query(Rson, qx);
}
inline void update_Range(int u, int v)
{
    while(top[u] ^ top[v])
    {
        if(deep[top[u]] < deep[top[v]]) swap(u, v);
        update(1, 1, N, id[top[u]], id[u], -1);
        u = fa[top[u]];
    }
    if(deep[u] < deep[v]) swap(u, v);
    update(1, 1, N, id[v], id[u], -1);
}
inline void init()
{
    cnt = 0;
    for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
    scanf("%d", &N);
    init();
    for(int i=1, u, v; i<N; i++)
    {
        scanf("%d%d", &u, &v);
        _add(u, v);
    }
    deep[0] = 0;
    dfs_1(1, 0);
    dfs_2(1, 1);
    scanf("%d", &Q);
    int op, x, tmp;
    while(Q--)
    {
        scanf("%d%d", &op, &x);
        if(op == 1)
        {
            update(1, 1, N, id[x], id[x] + siz[x] - 1, 1);
        }
        else if(op == 2)
        {
            update_Range(x, 1);
        }
        else
        {
            tmp = query(1, 1, N, id[x]);
            printf(~tmp ? "1\n" : "0\n");
        }
    }
    return 0;
}