CF911G Mass Change Queries 题解

合集 - codeforces题解集1(37)

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12.CF911G Mass Change Queries 题解2024-01-24

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题目链接：CF 或者 洛谷

前置知识点：平衡树合并: CF文章 与维基百科

看上去这题有很多人用线段树分裂与合并去做，其实这种需要分裂和合并的，我们用文艺平衡树去维护区间信息是最容易写的。

考虑本题的特殊性，值域并不是很大，所以其实我们可以为每种值开一棵文艺平衡树，而平衡树维护的值为下标序列，即对应的某个值的所有下标组成了一棵平衡树。而将某个范围内的值覆盖成另一个值，我们可以利用 FHQ 的分裂，传统性的拿到 $root[x]\mathrm{的}[l,r]$ 上的平衡树，然后合并到 $root[y]$ 对应的平衡树即可。

类似这题：梦幻布丁，我们也可以考虑启发式合并平衡树去做。不过更重要的是跟这题差不多的分析思想，考虑最坏的情况，就是 $[1,n]$ 上反复做操作，对 $[l,r]$ 上反复做操作也同理分析。很显然，每次操作要么不影响 $[l,r]$ 的颜色情况，要么减少一种颜色情况。不变的要么不存在这种颜色，要么这种颜色变成了另一种也不存在的颜色，这两种情况都是 $O(1)$ 的。考虑后者情况，很显然，它变成了梦幻布丁那题，同样的分析技巧容易知道，总操作上限复杂度不会超过：$n\log n$ 的，因为每次如果每次至少减少一种颜色，那么也最多减少了 $99$ 次，而它们的树节点总个数和为 $n$ 个节点。

最后，当然由于每个节点的值为下标标号，其实可以直接用节点下标表示，比如 $node[1]$ 既表示 $1$ 号 $FHQ$ 节点，也表示值，但为了可读性，还是采用传统的增加一个 $val$ 属性表示对应下标。最后就是 $dfs$ 所有树了，很显然，总节点个数不超过 $n$ 个，这个复杂度其实是 $O(n)$ 的。

参照代码

#include <bits/stdc++.h>
 
// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC optimize(2)
 
// #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 = 2e5 + 10;
constexpr int MX = 110;
 
struct FHQ
{
    int val, rnk, child[2];
    FHQ() = default;
 
    explicit FHQ(const int val)
        : val(val)
    {
        rnk = Rand(INT_MIN,INT_MAX);
        child[0] = child[1] = 0;
    }
} node[N];
 
#define rnk(x) node[x].rnk
#define val(x) node[x].val
#define left(x) node[x].child[0]
#define right(x) node[x].child[1]
int root[MX];
int cnt;
 
inline void split(const int curr, const int val, int& x, int& y)
{
    if (!curr)
    {
        x = y = 0;
        return;
    }
    if (val(curr) <= val)x = curr, split(right(curr), val,right(curr), y);
    else y = curr, split(left(curr), val, x,left(curr));
}
 
//合并左右两棵子树
inline int merge(const int x, const int y)
{
    if (!x or !y)return x ^ y;
    if (rnk(x) < rnk(y))
    {
        right(x) = merge(right(x), y);
        return x;
    }
    left(y) = merge(x,left(y));
    return y;
}
 
//从左到右合并三棵子树
inline int merge(const int l, const int mid, const int r)
{
    return merge(merge(l, mid), r);
}
 
//增加一个节点
inline void add(int& curr, const int val)
{
    int l = 0, r = 0;
    split(curr, val, l, r);
    node[++cnt] = FHQ(val);
    curr = merge(l, cnt, r);
}
 
//合并两棵平衡树
inline int mergeTree(int x, int y)
{
    if (!x or !y)return x ^ y;
    if (rnk(x) < rnk(y))swap(x, y);
    int leftTree, rightTree;
    split(y,val(x), leftTree, rightTree);
    left(x) = mergeTree(left(x), leftTree);
    right(x) = mergeTree(right(x), rightTree);
    return x;
}
 
 
int a[N];
int n, q;
int ans[N];
 
//dfs平衡树,如果有值就赋值
inline void dfs(const int curr, const int val)
{
    if (!curr)return;
    ans[val(curr)] = val;
    dfs(left(curr), val), dfs(right(curr), val);
}
 
inline void solve()
{
    cin >> n;
    forn(i, 1, n)cin >> a[i], add(root[a[i]], i);
    cin >> q;
    forn(i, 1, q)
    {
        int l, r, x, y;
        cin >> l >> r >> x >> y;
        int leftTree, rightTree, leftSon, rightSon; //[1,r]、[r+1,n]、[1,l-1]、[l,r]
        split(root[x], r, leftTree, rightTree);
        split(leftTree, l - 1, leftSon, rightSon);
        root[x] = mergeTree(leftSon, rightTree); //去掉[l,r]以后的平衡树合并
        root[y] = mergeTree(root[y], rightSon); //合并值y所在平衡树与[l,r]所在平衡树
    }
    forn(i, 1, 100)dfs(root[i], i);
    forn(i, 1, n)cout << ans[i] << ' ';
}
 
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;
}
 

$$\mathrm{时}\mathrm{间}\mathrm{复}\mathrm{杂}\mathrm{度}\mathrm{为}:O((n+q)\log n)$$