[AcWing 241] 楼兰图腾

树状数组 复杂度 $n \cdot log(n)$

单点修改， 区间查询

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

int n;
int a[N];
int tr[N];
int g[N], l[N];

int lowbit(int x)
{
    return x & -x;
}

void add(int x, int c)
{
    for (int i = x; i <= n; i += lowbit(i))
        tr[i] += c;
}

int ask(int x)
{
    int res = 0;
    for (int i = x; i; i -= lowbit(i))
        res += tr[i];
    return res;
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
        cin >> a[i];
    for (int i = 1; i <= n; i ++) {
        int y = a[i];
        g[i] = ask(n) - ask(y);
        l[i] = ask(y - 1);
        add(y, 1);
    }
    memset(tr, 0, sizeof tr);
    LL res1 = 0, res2 = 0;
    for (int i = n; i; i --) {
        int y = a[i];
        res1 += g[i] * 1LL * (ask(n) - ask(y));
        res2 += l[i] * 1LL * (ask(y - 1));
        add(y, 1);
    }
    cout << res1 << ' ' << res2 << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    solve();

    return 0;
}

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1. 对于 "V" 型，使用树状数组维护 $y_i$ 左边和右边比 $y_i$ 大的数的个数，另一种形状类似，使用树状数组维护 $y_i$ 左边和右边比 $y_i$ 小的数的个数