【CDQ分治】三元环

三元环

HDU - 7439

思路

考虑 \(3\) 个点的有向图,要么成环,要么有一个点入度为 \(2\) ,假设第 个点的入度为 \(d_i\),答案为 \(C_n^3-\sum\limits_{i=1}^nC_{d_i}^2\)

根据题目关系,\(i\rightarrow j\) 当且仅当 \(i<j \ and\ f_i <f_j \ and\ g_i < g_j\),否则就是 \(j\rightarrow i\),所以根据这个三维关系,我们可以先根据前两维求出 \(i<j\ and\ f_i\ge f_j\) 的入度,然后通过 cdq分治去求满足这个三维关系的各点的度数。

代码

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

template<typename T>
struct BIT {
#ifndef lowbit
#define lowbit(x) (x & (-x));
#endif
    int n;
    vector<T> t;

    BIT () {}
    BIT (int _n): n(_n) { t.resize(_n + 1); }
    BIT (int _n, vector<T>& a): n(_n) {
        t.resize(_n + 1);
        for (int i = 1; i <= n; ++ i) {
            t[i] += a[i];
            int j = i + lowbit(i);
            if (j <= n) t[j] += t[i];
        }
    }
    //单点修改
    void update(int i, T x) {
        while (i <= n) {
            t[i] += x;
            i += lowbit(i);
        }
    }
    //区间查询
    T sum(int i) {
        T ans = 0;
        while (i > 0) {
            ans += t[i];
            i -= lowbit(i);
        }
        return ans;
    }

    T query(int i, int j) {
        return sum(j) - sum(i - 1);
    }
    //区间修改则存入差分数组,[l, r] + k则update(x,k),update(y+1,-k)
    //单点查询则直接求前缀和sum(x)

    //求逆序对
    /*
    iota(d.begin(), d.end(), 0);
    stable_sort(d.begin(), d.end(), [&](int x, int y) {
        return a[x] < a[y];
    });去重排序

    BIT<i64> tree(n);
    i64 ans = 0;
    for (int i = 1; i <= n; i ++) {
        tree.update(d[i], 1);
        ans += i - tree.sum(d[i]);
    }
    */
};

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

    int n;
    cin >> n;

    vector<array<int, 3>> a(n + 1);
    for (int i = 1; i <= n; i ++) {
        cin >> a[i][0];
        a[i][2] = i;
    }

    for (int i = 1; i <= n; i ++) {
        cin >> a[i][1];
    }

    BIT<i64> bit(n);
    vector<int> in(n + 1);
    //求 i < j and fi >= fj
    for (int i = n; i >= 1; i --) {
        in[i] += bit.sum(a[i][0]);
        bit.update(a[i][0], 1);
    }

    for (int i = n; i >= 1; i --) {
        bit.update(a[i][0], -1);
    }

    auto cdq = [&](auto && self, int l, int r)->void{
        if (l == r)
            return ;

        int mid = l + r >> 1;
        self(self, l, mid);
        self(self, mid + 1, r);

        sort(a.begin() + l, a.begin() + mid + 1, [](auto x, auto y) {
            if (x[0] != y[0]) return x[0] < y[0];
            return x[1] < y[1];
        });

        sort(a.begin() + mid + 1, a.begin() + r + 1, [](auto x, auto y) {
            if (x[0] != y[0]) return x[0] < y[0];
            return x[1] < y[1];
        });

        //求 i < j and fi < fj and gi < gj
        int i = l, j = mid + 1;
        while (j <= r) {
            while (i <= mid && a[i][0] < a[j][0]) {
                bit.update(a[i][1], 1);
                i ++;
            }
            in[a[j][2]] += bit.sum(a[j][1] - 1);
            j ++;
        }
        for (int k = l; k < i; k ++) {
            bit.update(a[k][1], -1);
        }

        //求 i < j and fi < fj and gi >= gj
        i = mid, j = r;
        while (i >= l) {
            while (j > mid && a[j][0] > a[i][0]) {
                bit.update(a[j][1], 1);
                j --;
            }
            in[a[i][2]] += bit.sum(a[i][1]);
            i --;
        }
        for (int k = r; k > j; k --) {
            bit.update(a[k][1], -1);
        }
    };

    cdq(cdq, 1, n);

    i64 ans = 1ll * n * (n - 1) * (n - 2) / 6;
    for (int i = 1; i <= n; i ++) {
        ans -= 1ll * in[i] * (in[i] - 1) / 2;
    }

    cout << ans << '\n';

    return 0;
}
posted @ 2024-08-06 20:45  Ke_scholar  阅读(3)  评论(0编辑  收藏  举报