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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF785E Anton and Permutation

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CF785E Anton and Permutation

CF: E. Anton and Permutation

Algorithm I 线段树套平衡树

比较裸，大常数两只log，过不了。

查看代码 
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#include <cstdio>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
typedef long long LL;
typedef tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> Tree;
const int N = 2e5 + 10;
LL ans;
int n, m, w[N];
struct Node { int l, r; Tree t; } tr[N << 2];
void build (int l = 1, int r = n, int x = 1)
{
    tr[x].l = l, tr[x].r = r;
    for (int i = l; i <= r; ++i) tr[x].t.insert(w[i]);
    if (l == r) return;
    int mid = l + r >> 1;
    build(l, mid, x << 1), build(mid + 1, r, x << 1 | 1);
}
void add (int t, int x = 1)
{
    tr[x].t.insert(w[t]);
    if (tr[x].l == tr[x].r) return;
    int mid = tr[x].l + tr[x].r >> 1;
    add(t, x << 1 | (t > mid));
}
void del (int t, int x = 1)
{
    tr[x].t.erase(w[t]);
    if (tr[x].l == tr[x].r) return;
    int mid = tr[x].l + tr[x].r >> 1;
    del(t, x << 1 | (t > mid));
}
int QueryLess (int l, int r, int k, int x = 1)
{
    if (l > r || tr[x].l > r || l > tr[x].r) return 0;
    if (tr[x].l >= l && tr[x].r <= r) return tr[x].t.order_of_key(k);
    return QueryLess(l, r, k, x << 1) + QueryLess(l, r, k, x << 1 | 1);
}
int QueryGreater (int l, int r, int k, int x = 1)
{
    if (l > r || tr[x].l > r || l > tr[x].r) return 0;
    if (tr[x].l >= l && tr[x].r <= r) return tr[x].t.size() - tr[x].t.order_of_key(k + 1);
    return QueryGreater(l, r, k, x << 1) + QueryGreater(l, r, k, x << 1 | 1);
}
int main ()
{
    read(n, m);
    for (int i = 1; i <= n; ++i) w[i] = i;
    build();
    for (int a, b; m; --m)
    {
        read(a, b);
        if (a > b) swap(a, b);
        ans -= QueryLess(a + 1, b - 1, w[a]) + QueryGreater(a + 1, b - 1, w[b]) + (w[a] > w[b]);
        ans += QueryLess(a + 1, b - 1, w[b]) + QueryGreater(a + 1, b - 1, w[a]) + (w[b] > w[a]);
        del(a), del(b), swap(w[a], w[b]), add(a), add(b);
        write(ans), puts("");
    }
    return 0;
}

Algorithm II 分块套树状数组

在 OI-Wiki 上被称为这个名称，并被分入“树套树”标题下，实际上问题就是二维偏序，如果直接用树状数组那么空间爆炸。分块后空间由 $O(n ^ 2)$ 减少为 $O(n ^ {\frac 3 2})$ ，可以接受。复杂度 单次复杂度 $O(\sqrt n + log ^ 2)$ ，常数较小。

查看代码 
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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar('0' + x % 10);
}
typedef long long LL;
const int N = 2e5 + 10, B = 450;
LL ans;
int n, m, w[N], tot, id[N], L[N], R[N], tr[B][N];
void add (int x, int y, int k)
{
    for (int i = x; i <= tot; i += i & -i)
        for (int j = y; j <= n; j += j & -j)
            tr[i][j] += k;
}
int query (int x, int y)
{
    int res = 0;
    for (int i = x; i; i -= i & -i)
        for (int j = y; j; j -= j & -j)
            res += tr[i][j];
    return res;
}
int QueryLess (int l, int r, int k) { return query(r, k - 1) - query(l - 1, k - 1); }
int QueryGreater (int l, int r, int k) { return (r - l + 1) - QueryLess(l, r, k + 1); }
int main ()
{
    read(n, m);
    for (int i = 1; i <= n; ++i)
        id[i] = (i - 1) / B + 1;
    tot = id[n];
    for (int i = 1; i <= tot; ++i)
        L[i] = R[i - 1] + 1, R[i] = R[i - 1] + B;
    R[tot] = n;
    for (int i = 1; i <= n; ++i) add(id[i], w[i] = i, 1);
    for (int a, b; m; --m)
    {
        read(a, b);
        if (a > b) swap(a, b);
        if (id[a] == id[b])
        {
            ans += (w[b] > w[a]) - (w[a] > w[b]);
            for (int i = a + 1; i < b; ++i)
                ans += (w[b] > w[i]) + (w[i] > w[a]) - (w[a] > w[i]) - (w[i] > w[b]);
            swap(w[a], w[b]);
        }
        else
        {
            ans += (w[b] > w[a]) - (w[a] > w[b]);
            for (int i = a + 1; i <= R[id[a]]; ++i)
                ans += (w[b] > w[i]) + (w[i] > w[a]) - (w[a] > w[i]) - (w[i] > w[b]);
            ans -= QueryGreater(id[a] + 1, id[b] - 1, w[b]) + QueryLess(id[a] + 1, id[b] - 1, w[a]);
            ans += QueryGreater(id[a] + 1, id[b] - 1, w[a]) + QueryLess(id[a] + 1, id[b] - 1, w[b]);
            for (int i = L[id[b]]; i < b; ++i)
                ans += (w[b] > w[i]) + (w[i] > w[a]) - (w[a] > w[i]) - (w[i] > w[b]);
            add(id[a], w[a], -1), add(id[b], w[b], -1);
            swap(w[a], w[b]);
            add(id[a], w[a], 1), add(id[b], w[b], 1);
        }
        write(ans), puts("");
    }
    return 0;
}
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