CF459D Pashmak and Parmida's problem

题意

给定一个 $n$ 个元素的序列 $a$，设 $f(i, j, x)$ 表示 $a_i \sim a_j$ 中 $x$ 出现次数，求有多少对 $i, j$ 满足 $i < j$ 且 $f(1, i, a_i) > f(j, n, a_j)$。

解法

令 $p_i =f(1, i, a_i), g_i = f(i, n, a_i)$，那么我们可以将这些操作转化成逆序对问题。

$O(n)$ 处理 $p, g$ 序列，$O(n \log n)$ 值域线段树维护逆序对个数即可。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

#define int unsigned long long

const int N = 1e6 + 5;

int n, a[N], cnta[N], cntb[N];
int ca[N], cb[N];
vector<int> b;

class SegmentTree
{
public:
	struct Node
	{
		int l, r, sum;
	}tr[N << 2];
	void push_up(int u)
	{
		tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
	}
	void build(int u, int l, int r)
	{
		tr[u] = { l, r, 0 };
		if (l == r) return;
		int mid = l + r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
	}
	void update(int u, int x)
	{
		if (tr[u].l == tr[u].r)
		{
			tr[u].sum++;
			return;
		}
		int mid = tr[u].l + tr[u].r >> 1;
		if (x <= mid) update(u << 1, x);
		else update(u << 1 | 1, x);
		push_up(u);
	}
	int query(int u, int l, int r)
	{
		if (tr[u].l >= l and tr[u].r <= r) return tr[u].sum;
		int mid = tr[u].l + tr[u].r >> 1, res = 0;
		if (l <= mid) res = query(u << 1, l, r);
		if (r > mid) res += query(u << 1 | 1, l, r);
		return res;
	}
};

SegmentTree seg;

int main()
{
	scanf("%llu", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%llu", &a[i]);
		b.push_back(a[i]);
	}
	sort(b.begin(), b.end());
	b.erase(unique(b.begin(), b.end()), b.end());
	for (int i = 1; i <= n; i++) a[i] = lower_bound(b.begin(), b.end(), a[i]) - b.begin() + 1;
	for (int i = 1; i <= n; i++)
	{
		ca[a[i]]++;
		cnta[i] = ca[a[i]];
	}
	for (int i = n; i >= 1; i--)
	{
		cb[a[i]]++;
		cntb[i] = cb[a[i]];
	}
	int ans = 0;
	seg.build(1, 1, n);
	for (int i = n; i >= 1; i--)
	{
		ans += seg.query(1, 1, cnta[i] - 1);
		seg.update(1, cntb[i]);
	}
	printf("%llu\n", ans);
	return 0;
}