Bzoj2141: 排队

题面

传送门

Sol

树状数组套线段树模板题

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(2e4 + 1);
typedef long long ll;

IL int Input(){
	RG int x = 0, z = 1; RG char c = getchar();
	for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
	for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x * z;
}

int h[_], n, num, rt[_], o[_], len;
struct Segment{
	int ls, rs, sz;
} T[_ * 200];
ll ans;

IL void Modify(RG int &x, RG int l, RG int r, RG int p, RG int v){
	if(!x) x = ++num; T[x].sz += v;
	if(l == r) return;
	RG int mid = (l + r) >> 1;
	if(p <= mid) Modify(T[x].ls, l, mid, p, v);
	else Modify(T[x].rs, mid + 1, r, p, v);
}

IL void Add(RG int x, RG int v, RG int d){
	for(; x <= n; x += x & -x) Modify(rt[x], 1, len, v, d);
}

IL int QuerySum(RG int x, RG int l, RG int r, RG int ql, RG int qr){
	if(!x) return 0;
	if(ql <= l && qr >= r) return T[x].sz;
	RG int mid = (l + r) >> 1, ret = 0;
	if(ql <= mid) ret = QuerySum(T[x].ls, l, mid, ql, qr);
	if(qr > mid) ret += QuerySum(T[x].rs, mid + 1, r, ql, qr);
	return ret;
}

IL int Sum(RG int x, RG int l, RG int r){
	RG int ret = 0;
	if(l > r) return 0;
	for(; x; x -= x & -x) ret += QuerySum(rt[x], 1, len, l, r);
	return ret;
}

int main(RG int argc, RG char* argv[]){
	n = Input();
	for(RG int i = 1; i <= n; ++i) o[++len] = h[i] = Input();
	sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1;
	for(RG int i = 1; i <= n; ++i){
		h[i] = lower_bound(o + 1, o + len + 1, h[i]) - o;
		ans += Sum(i, h[i] + 1, len), Add(i, h[i], 1);
	}
	printf("%lld\n", ans);
	for(RG int m = Input(); m; --m){
		RG int x = Input(), y = Input();
		if(x > y) swap(x, y);
		Add(x, h[x], -1), Add(y, h[y], -1);
		ans -= Sum(x, h[x] + 1, len) + Sum(y, h[y] + 1, len);
		ans -= Sum(n, 1, h[x] - 1) - Sum(x - 1, 1, h[x] - 1);
		ans -= Sum(n, 1, h[y] - 1) - Sum(y - 1, 1, h[y] - 1);
		if(h[x] > h[y]) --ans;
		swap(h[x], h[y]);
		if(h[x] > h[y]) ++ans;
		ans += Sum(x, h[x] + 1, len) + Sum(y, h[y] + 1, len);
		ans += Sum(n, 1, h[x] - 1) - Sum(x - 1, 1, h[x] - 1);
		ans += Sum(n, 1, h[y] - 1) - Sum(y - 1, 1, h[y] - 1);
		Add(x, h[x], 1), Add(y, h[y], 1);
		printf("%lld\n", ans);
	}
    return 0;
}