2019陕西省赛-E Tree

Tree

树链剖分 + 线段树

除去区间修改的难点，其他都是树链剖分基础

线段树维护异或区间和，用且、或修改区间单点，如果单考虑一个值的话似乎很难修改，因此考虑将一个数字分割成 32 位，硬跑 32 次

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 1e5 + 10;
int tr[35][maxn << 2], lazy[35][maxn << 2], num[maxn];
int dep[maxn], siz[maxn], hson[maxn], fa[maxn];
int dfn[maxn], rnk[maxn], top[maxn];
vector<int>gra[maxn];

void dfs1(int now, int pre, int d)
{
    dep[now] = d;
    hson[now] = -1;
    fa[now] = pre;
    siz[now] = 1;
    for(auto nex : gra[now])
    {
        if(nex == pre) continue;
        dfs1(nex, now, d + 1);
        siz[now] += siz[nex];
        if(hson[now] == -1 || siz[hson[now]] < siz[nex])
            hson[now] = nex;
    }
}

int tp = 0;
void dfs2(int now, int t)
{
    top[now] = t;
    tp++;
    dfn[now] = tp;
    rnk[tp] = now;
    if(hson[now] != -1)
    {
        dfs2(hson[now], t);
        for(auto nex : gra[now])
        {
            if(nex == hson[now] || nex == fa[now]) continue;
            dfs2(nex, nex);
        }
    }
}

void build(int now, int l, int r)
{
    if(l == r)
    {
        int nex = rnk[l];
        for(int i=0; i<32 && num[nex]; i++)
        {
            tr[i][now] = num[nex] & 1;
            num[nex] >>= 1;
        }
        return;
    }
    int mid = l + r >> 1;
    build(now << 1, l, mid);
    build(now << 1 | 1, mid + 1, r);
    for(int i=0; i<32; i++) tr[i][now] = tr[i][now << 1] + tr[i][now << 1 | 1];
}

int bit = 0;
inline void push_down(int now, int l, int r)
{
    if(lazy[bit][now] == 2)
    {
        tr[bit][now << 1] = tr[bit][now << 1 | 1] = 0;
        lazy[bit][now << 1] = lazy[bit][now << 1 | 1] = 2;
    }
    else if(lazy[bit][now] == 1)
    {
        int mid = l + r >> 1;
        tr[bit][now << 1] = mid - l + 1;
        tr[bit][now << 1 | 1] = r - mid;
        lazy[bit][now << 1] = lazy[bit][now << 1 | 1] = 1;
    }
    lazy[bit][now] = 0;
}

void update(int now, int l, int r, int L, int R, int t)
{
    if(L <= l && r <= R)
    {
        if(t == 2) tr[bit][now] = 0;
        else tr[bit][now] = r - l + 1;
        lazy[bit][now] = t;
        return;
    }
    push_down(now, l, r);
    int mid = l + r >> 1;
    if(L <= mid)
        update(now << 1, l, mid, L, R, t);
    if(R > mid)
        update(now << 1 | 1, mid + 1, r, L, R, t);
    tr[bit][now] = tr[bit][now << 1] + tr[bit][now << 1 | 1];
}

int query(int now, int l, int r, int L, int R)
{
    if(L <= l && r <= R)
        return tr[bit][now];
    push_down(now, l, r);
    int mid = l + r >> 1, ans = 0;
    if(L <= mid)
        ans += query(now << 1, l, mid, L, R);
    if(R > mid)
        ans += query(now << 1 | 1, mid + 1, r, L, R);
    return ans;
}

void init(int n, int rt = 1)
{
    tp = 0;
    dfs1(rt, rt, 1);
    dfs2(rt, rt);
    build(1, 1, n);
}

bool solve(int n, int a, int t)
{
    bit = 0;
    for(int i=0; i<32; i++, bit++)
    {
        int now = a, b = t & 1;
        t >>= 1;
        while(top[now] != 1)
        {
            b += query(1, 1, n, dfn[top[now]], dfn[now]);
            now = fa[top[now]];
        }
        b += query(1, 1, n, 1, dfn[now]);
        if(b & 1) return true; 
    }
    return false;
}

void update_a(int n, int s, int t, int op)
{
    int way = 1;
    if(op == 2) way = 0;
    bit = 0;
    for(; bit < 32; t >>= 1, bit++)
    {
        if(way != (t & 1)) continue;
        int a = s;
        while(top[a] != 1)
        {
            update(1, 1, n, dfn[top[a]], dfn[a], op);
            a = fa[top[a]];
        }
        update(1, 1, n, 1, dfn[a], op);
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++) scanf("%d", &num[i]);
    for(int i=1; i<n; i++)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        gra[y].push_back(x);
        gra[x].push_back(y);
    }
    init(n);
    while(m--)
    {
        int op, s, t;
        scanf("%d%d%d", &op, &s, &t);
        if(op == 3)
        {
            if(solve(n, s, t))
                printf("YES\n");
            else
                printf("NO\n");
        }
        else
            update_a(n, s, t, op);
    }
    return 0;
}