Luogu P3157 [CQOI2011]动态逆序对

题目链接 \(Click\) \(Here\)

这个题有点卡常数。。我的常数比较大所以是吸着氧气跑过去的。。。

题意：计算对于序列中每个位置\(p\)，\([1,p-1]\)区间内比它大的数的个数，和\([p + 1, N]\)区间内比它小的数的个数和，要求支持修改操作，带修主席树可以解决。

通过主席树来维护权值状态和比某个数大/小的数的个数，用树状数组来支持修改和维护一个主席树的前缀和（主席树前缀和具有可减性）。时间空间\(O(Nlog^2N)\)，\(1000ms＋128MB\)的限制对本算法略为苛刻，但卡常可过。

#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
#define mid ((l + r) >> 1)
#define lowbit(x) (x & -x)
int n, m, tot, rt[N], arr[N], pos[N];
struct Segment_Node {
	int sz, ls, rs;
}t[N << 8];
void modify (int &_rt, int l, int r, int w, int del) {
	if (_rt == 0) _rt = ++tot;
	t[_rt].sz += del;
	if (l != r) {
		if (w <= mid) {
			modify (t[_rt].ls, l, mid, w, del);
		} else {
			modify (t[_rt].rs, mid + 1, r, w, del);
		}
	}
}
namespace rev {
	int a[N];
	inline void add (int pos, int val) {
		while (pos <= n) {
			a[pos] += val;
			pos += lowbit (pos);
		}
	}
	inline int get_sum (int pos) {
		register int res = 0;
		while (pos) {
			res += a[pos];
			pos -= lowbit (pos);
		}
		return res;
	}
	long long get_rev () {
		long long res = 0;
		for (int i = n; i >= 1;--i) {
			res += get_sum (arr[i]);
			add (arr[i], 1);
		}
		return res;
	}
}
int _query (int _rt, int l, int r, int nl, int nr) {
	if (nl <= l && r <= nr) return t[_rt].sz;
	register int res = 0;
	if (nl <= mid) res += _query (t[_rt].ls, l, mid, nl, nr);
	if (mid < nr) res += _query (t[_rt].rs, mid + 1, r, nl, nr);
	return res;
}
inline int query (int l, int r, int w, int type) {
    l = l - 1;
	//求序列里面[l, r]内有多少数大于w (type = 1)
	//求序列里面[l, r]内有多少数小于w (type = 2)
	// printf ("l = %d, r = %d, w = %d, type = %d\n", l, r, w, type);
	register int i, res = 0;
	for (i = l; i != 0; i -= lowbit (i)) {
		if (type == 1) res -= _query (rt[i], 0, n + 1, w + 1, n);
		if (type == 2) res -= _query (rt[i], 0, n + 1, 1, w - 1);
	}
	// printf ("res = %d\n", res);
	for (i = r; i != 0; i -= lowbit (i)) {
		if (type == 1) res += _query (rt[i], 0, n + 1, w + 1, n);
		if (type == 2) res += _query (rt[i], 0, n + 1, 1, w - 1);
	}
	// printf ("l = %d, r = %d, w = %d, type = %d, res = %d\n", l, r, w, type, res);
	return res;
}
inline int read () {
	int s = 0, w = 1, ch = getchar ();
	while ('9' < ch || ch < '0') {
		ch = getchar ();
	}
	while ('0' <= ch && ch <= '9') {
		s = s * 10 + ch - '0';
		ch = getchar ();
	}
	return s * w;
}
int main () {
	n = read (), m = read ();
	register int i, j, w;
	for (i = 1; i <= n; ++i) {
	    arr[i] = read ();
		pos[arr[i]] = i;
		for (j = i; j <= n; j += lowbit (j)) {
			modify (rt[j], 0, n + 1, arr[i], +1);
		}
	}
	long long ans = rev :: get_rev ();
	for (i = 1, w = 0; i <= m; ++i) {
		printf ("%lld\n", ans);
		w = read ();
		ans -= query (1, pos[w] - 1, w, 1);
		ans -= query (pos[w] + 1, n, w, 2);
		for (j = pos[w]; j <= n; j += lowbit (j)) {
			modify (rt[j], 0, n + 1, w, -1);
		}
	}
}