CF620E New Year Tree 题解

题目链接：CF 或者 洛谷

这题很简单，看到颜色数，HH的项链？树，树上的HH的项链？带修，树上的镜中的昆虫？\(c_i \le 60\)，噢，easy 了。

考虑一类信息，表示有和无，对于某种颜色来讲，\(0/1\) 表示无或者有，而或运算让我们从小区间的有无情况反映到大区间的有无情况。一种暴力的想法，为每种颜色建立一棵有无颜色的线段树，每次依次查询这 \(60\) 棵线段树对应区间内的该颜色的有无情况，\(60\times m\log{n}\) 看上去可行，不过我们有一种更简单的方式反应一堆量的有无，显然是：\(状态压缩\)，观察到颜色不超过 \(60\) 种，我们每个维护一个 \(64\ 位的\ val\) 即可。观察到有覆盖操作，那么再加一个覆盖懒标记就好了。由于跟子树有关，那就别考虑树剖了，直接 \(dfs\) 序简简单单的，每次我们可以用 \(1<<(val-1)\) 表示 \(val\) 表示这种颜色是有的。线段树维护状态压缩即可。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

// #define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 4e5 + 10;

struct Node
{
    ll val, cover;
} node[N << 2];

#define val(x) node[x].val
#define cover(x) node[x].cover

inline void push_up(const int curr)
{
    val(curr) = val(ls(curr)) | val(rs(curr));
}

inline void push_down(const int curr)
{
    if (cover(curr))
    {
        val(ls(curr)) = val(rs(curr)) = cover(ls(curr)) = cover(rs(curr)) = cover(curr);
        cover(curr) = 0;
    }
}

int n, q;

inline void Cover(const int curr, const int l, const int r, const ll val, const int s = 1, const int e = n)
{
    const int mid = s + e >> 1;
    if (l <= s and e <= r)
    {
        val(curr) = cover(curr) = val;
        return;
    }
    push_down(curr);
    if (l <= mid)Cover(ls(curr), l, r, val, s, mid);
    if (r > mid)Cover(rs(curr), l, r, val, mid + 1, e);
    push_up(curr);
}

inline ll Query(const int curr, const int l, const int r, const int s = 1, const int e = n)
{
    if (l <= s and e <= r)return val(curr);
    push_down(curr);
    const int mid = s + e >> 1;
    ll ans = 0;
    if (l <= mid)ans |= Query(ls(curr), l, r, s, mid);
    if (r > mid)ans |= Query(rs(curr), l, r, mid + 1, e);
    return ans;
}

int cnt, s[N], e[N];
vector<int> child[N];
int a[N];
int rnk[N];
inline void dfs(const int curr, const int fa)
{
    s[curr] = ++cnt;
    rnk[cnt]=curr;
    for (const auto nxt : child[curr])if (nxt != fa)dfs(nxt, curr);
    e[curr] = cnt;
}

inline void build(const int curr = 1, const int l = 1, const int r = n)
{
    const int mid = l + r >> 1;
    if (l == r)
    {
        val(curr) = 1LL << a[rnk[l]];
        return;
    }
    build(ls(curr), l, mid);
    build(rs(curr), mid + 1, r);
    push_up(curr);
}

inline void solve()
{
    cin >> n >> q;
    forn(i, 1, n)cin >> a[i], a[i]--;
    forn(i, 1, n-1)
    {
        int u, v;
        cin >> u >> v;
        child[u].push_back(v), child[v].push_back(u);
    }
    dfs(1, 0);
    build();
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int u, val;
            cin >> u >> val;
            Cover(1, s[u], e[u], 1LL << --val);
        }
        else
        {
            int u;
            cin >> u;
            ll ans = Query(1, s[u], e[u]);
            cout << __popcount(ans) << endl;
        }
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度忽略掉\ popcount\ 的复杂度：O(t\log{n}) \]