2022年“腾讯杯”大学生程序设计竞赛 死去的 Elo 突然开始攻击我 题解

题目链接：死去的 Elo 突然开始攻击我

容易知道，如果暴力对某个区间而言进行查询，我们可以考虑使用并查集，开一个桶，每次添加一个数 \(val\)，那么如果已经存在了 \(val-1\) 或者 \(val+1\) ，我们可以考虑合并 \(val\) 与 \(val-1\) 或者 \(val+1\) 在同一个集合中，查询答案就是最大的连通分量。

考虑莫队做法，如何并查集比较容易插入，但并不容易删除，但非常好撤销！所以我们考虑回滚莫队。具体原因可以详情见我的另外两篇文章，最后给出。那么就非常简单了，回滚莫队使用只加不减，在左端点回滚时撤销原来的 \(merge\) 即可。对于暴力我们可以用一个可以支持路径压缩的普通并查集即可，非暴力使用启发式合并的并查集保证复杂度。常数很小：\(O(\log{\sqrt{n}})\)。

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 = 5e4 + 10;
int fa[N], siz[N], curr = 1; //需要回滚的
stack<tuple<int, int, int, int>> st; //x,y,fa[x],siz[y],preCurr
int tmpFa[N], tmpSiz[N], tmpCurr = 1; //不需要回滚的
stack<int> Cnt;
//不能路径压缩
inline int find(const int x)
{
    return x == fa[x] ? x : find(fa[x]);
}

//可以路径压缩
inline int tmpFind(const int x)
{
    return x == tmpFa[x] ? x : tmpFa[x] = tmpFind(tmpFa[x]);
}

//需要回滚记得用栈记录信息
inline void merge(int x, int y, const bool isNeedBack)
{
    x = find(x), y = find(y);
    if (siz[x] > siz[y])swap(x, y);
    if (isNeedBack)st.emplace(x, y, fa[x], siz[y]);
    fa[x] = y, siz[y] += siz[x], uMax(curr, siz[y]);
}

//直接合并
inline void tmpMerge(int x, int y)
{
    x = tmpFind(x), y = tmpFind(y);
    tmpFa[x] = y;
    tmpSiz[y] += tmpSiz[x];
    uMax(tmpCurr, tmpSiz[y]);
}

int pos[N]; //序列分块
//回滚莫队按照右端点升序
struct Mo
{
    int l, r, id;

    bool operator<(const Mo& other) const
    {
        return pos[l] ^ pos[other.l] ? l < other.l : r < other.r;
    }
} node[N];

int cnt[N], tmpCnt[N]; //两种桶,暴力用的和非暴力的
int n, q, a[N];
int ans[N]; //答案
int e[N]; //块的右端点
//是否需要回滚的插入
inline void add(const int val, const bool isNeedBack)
{
    cnt[val]++;
    if (cnt[val - 1])merge(val - 1, val, isNeedBack);
    if (cnt[val + 1])merge(val + 1, val, isNeedBack);
    if (isNeedBack)Cnt.push(val);
}

//暴力插入
inline void tmpAdd(const int val)
{
    tmpCnt[val]++;
    if (tmpCnt[val - 1])tmpMerge(val - 1, val);
    if (tmpCnt[val + 1])tmpMerge(val + 1, val);
}

inline void solve()
{
    cin >> n >> q;
    int blockSize = sqrt(n);
    int blockCnt = (n + blockSize - 1) / blockSize;
    forn(i, 1, n)cin >> a[i], pos[i] = (i - 1) / blockSize + 1;
    forn(i, 1, n)siz[i] = tmpSiz[i] = 1, fa[i] = tmpFa[i] = i;
    forn(i, 1, blockCnt)e[i] = i * blockSize;
    e[blockCnt] = n;
    forn(i, 1, q)
    {
        auto& [l,r,id] = node[i];
        cin >> l >> r, id = i;
    }
    sortArr(node, q);
    int l = 1, r = 0, last = 0;
    forn(i, 1, q)
    {
        auto [L,R,id] = node[i];
        if (pos[L] == pos[R])
        {
            forn(idx, L, R)tmpAdd(a[idx]);
            ans[id] = tmpCurr;
            tmpCurr = 1;
            forn(idx, L, R)tmpSiz[a[idx]] = 1, tmpCnt[a[idx]] = 0, tmpFa[a[idx]] = a[idx]; //记得恢复
            continue;
        }
        //和上一次左端点查询的块不同,更新新块查询
        if (last != pos[L])
        {
            forn(idx, 1, n)siz[idx] = 1, fa[idx] = idx, cnt[idx] = 0;
            curr = 1;
            l = e[pos[L]] + 1;
            r = l - 1;
            last = pos[L];
        }
        while (r < R)add(a[++r], false); //右端点不需要回滚
        int tmpL = l;
        int preCurr = curr; //回滚答案
        while (tmpL > L)add(a[--tmpL], true); //左端点需要回滚
        ans[id] = curr;
        //回滚信息
        while (!st.empty())
        {
            auto [x,y,faX,sizY] = st.top();
            st.pop();
            fa[x] = faX, siz[y] = sizY;
        }
        curr = preCurr;
        while (!Cnt.empty())
        {
            auto val = Cnt.top();
            Cnt.pop();
            cnt[val]--;
        }
    }
    forn(i, 1, q)cout << ans[i] << 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;
}

\[时间复杂度为 \ O(\log{\sqrt{n}} \times q\sqrt{n}),当然也可以用链表代替可撤销并查集优化掉常数。
\]

关于我的回滚莫队讲解的两个常见题：

例题1 核心思想讲解

例题2 和本题几乎一致的题型