P1975 [国家集训队] 排队 题解

题目链接：排队

水紫，\(n\) 不大，树套树或者分块都能做。分块的话，最优序列分块套套值域分块最优。观察到是可差性问题维护，即权值数量维护，那我们就 树状数组套权值线段树 即可。由于 \(n\) 不大，我们可以不用回收标记，直接数组空间开大点就行。我们预处理出初始逆序对，每一次操作都是基于它的差分变化，讨论变化即可。预处理树套树或者普通 bit 都行。变化，先保证下标为 \(i\le j\)，不满足就 \(swap\) 一下，讨论下，如果 \([i,j]\) 之间有数，当做 \([l+1,r-1]\) 之间有数，我们考虑一次交换操作的贡献变化，计算出分别大于小于 \(l\) 处数的数数量，和 \(r\) 处的，很显然，\(l\) 处 \([l+1,r-1]\) 比 \(a[l]\) 小的数贡献没了，多了比它大的数，\(r\) 处则相反。最后考虑 \(a[l]\) 和 \(a[r]\) 的贡献变化就行了，顺便更新树套树。

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 = 1e5 + 10;
constexpr int MX = 1e9 + 1;
int n, q;

struct
{
    int tot;

    struct In
    {
        int left, right, cnt;
    } in[N << 6];

#define left(x) in[x].left
#define right(x) in[x].right
#define cnt(x) in[x].cnt

    void add(int& curr, const int pos, const int l = 0, const int r = MX)
    {
        if (!curr)curr = ++tot;
        cnt(curr) += pos > 0 ? 1 : -1;
        const int mid = l + r >> 1;
        if (l == r)return;
        if (abs(pos) <= mid)add(left(curr), pos, l, mid);
        else add(right(curr), pos, mid + 1, r);
    }

    int query(const int curr, const int l, const int r, const int s = 0, const int e = MX)
    {
        if (!curr)return 0;
        const int mid = s + e >> 1;
        if (l <= s and e <= r)return cnt(curr);
        int ans = 0;
        if (l <= mid)ans += query(left(curr), l, r, s, mid);
        if (r > mid)ans += query(right(curr), l, r, mid + 1, e);
        return ans;
    }

    int root[N];

    void Add(int x, const int val)
    {
        while (x <= n)add(root[x], val), x += lowBit(x);
    }

    int Query(int x, const int l, const int r)
    {
        int ans = 0;
        while (x)ans += query(root[x], l, r), x -= lowBit(x);
        return ans;
    }

    int QueryLR(const int l, const int r, const int L, const int R)
    {
        return Query(r, L, R) - Query(l - 1, L, R);
    }
} seg;

ll ans;
int a[N];;

inline void solve()
{
    cin >> n;
    forn(i, 1, n)
    {
        cin >> a[i];
        ans += seg.Query(i - 1, a[i] + 1, MX);
        seg.Add(i, a[i]);
    }
    cout << ans << endl;
    cin >> q;
    while (q--)
    {
        int l, r;
        cin >> l >> r;
        if (l > r)swap(l, r);
        if (r - l > 1)
        {
            const int mxL = seg.QueryLR(l + 1, r - 1, a[l] + 1, MX);
            const int mxR = seg.QueryLR(l + 1, r - 1, a[r] + 1, MX);
            const int miL = seg.QueryLR(l + 1, r - 1, 0, a[l] - 1);
            const int miR = seg.QueryLR(l + 1, r - 1, 0, a[r] - 1);
            ans += mxL - miL + miR - mxR;
        }
        seg.Add(l, -a[l]), seg.Add(r, a[l]);
        seg.Add(r, -a[r]), seg.Add(l, a[r]);
        ans -= a[l] > a[r];
        ans += a[l] < a[r];
        swap(a[l], a[r]);
        cout << 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;
}

\[时间复杂度为:\ O((n+q)\log^2{V_{max}}) \]