[CF1830E] Bully Sort

题目描述

On a permutation $ p $ of length $ n $ , we define a bully swap as follows:

• Let $ i $ be the index of the largest element $ p_i $ such that $ p_i \neq i $ .
• Let $ j $ be the index of the smallest element $ p_j $ such that $ i < j $ .
• Swap $ p_i $ and $ p_j $ .

We define $ f(p) $ as the number of bully swaps we need to perform until $ p $ becomes sorted. Note that if $ p $ is the identity permutation, $ f(p)=0 $ .

You are given $ n $ and a permutation $ p $ of length $ n $ . You need to process the following $ q $ updates.

In each update, you are given two integers $ x $ and $ y $ . You will swap $ p_x $ and $ p_y $ and then find the value of $ f(p) $ .

Note that the updates are persistent. Changes made to the permutation $ p $ will apply when processing future updates.

输入格式

The first line of the input contains two integers $ n $ and $ q $ ( $ 2 \le n \le 5 \cdot 10^5 $ , $ 1 \le q \le 5 \cdot 10^4 $ ) — the length of the permutation and the number of updates.

The second line of input contains $ n $ integer $ p_1,p_2,\ldots,p_n $ ( $ 1 \leq p_i \leq n $ ) — the permutation $ p $ . All elements of $ p $ are distinct.

The $ i $ -th of the next $ q $ lines of input contains two integers $ x_i $ and $ y_i $ ( $ 1 \le x_i < y_i \le n $ ) — describing the $ i $ -th update.

神仙题。

现在主要问题是如何快速求出 \(f(p)\) 的答案。

结论是 \(\sum\limits_{i=1}^n |i-p_i|-\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n[a_i>a_j]\)

因为在一次交换 \(i,j\) 时，逆序对减少 \(2(j-i)-1\) 次，\(|i-p_i|\) 减少 \(2(j-i)\) 个，于是减一下就是答案。

然后动态维护逆序对就好，我打了个常数巨大的 KDT。

#include<bits/stdc++.h>
using namespace std;
const int N=6e5+5;
typedef long long LL;
LL ans;
int n,q,mxx[N],mnx[N],mxy[N],mny[N],idx,c[N],tr[N],g[N],p[N],m;
pair<int,int>a[N];
int cmp(pair<int,int>x,pair<int,int>y)
{
	return x.second^y.second? x.second<y.second:x.first<y.first;
}
int ask(int l,int r,int u,int d)
{
	int o=l+r>>1;
	if(l>r||mnx[o]>u||mxy[o]<d)
		return 0;
	if(mxx[o]<=u&&mny[o]>=d)
		return c[o];
	return ask(l,o-1,u,d)+ask(o+1,r,u,d)+(a[o].first<=u&&a[o].second>=d&&g[o]);
}
int build(int l,int r,int op)
{
	if(l>r)
		return 0;
	int o=l+r>>1;
	if(!op)
		nth_element(a+l,a+o,a+r+1);
	else
		nth_element(a+l,a+o,a+r+1,cmp);
	int lc=build(l,o-1,!op),rc=build(o+1,r,!op);
	mxx[o]=max({mxx[lc],mxx[rc],a[o].first});
	mnx[o]=min({mnx[lc],mnx[rc],a[o].first});
	mxy[o]=max({mxy[lc],mxy[rc],a[o].second});
	mny[o]=min({mny[lc],mny[rc],a[o].second});
	return o;
}
void add(int l,int r,int a0,int a1,int t,int op)
{
	int o=l+r>>1;
	c[o]+=t;
	if(a0==a[o].first&&a1==a[o].second)
		return g[o]+=t,void();
	if(!op&&make_pair(a0,a1)<a[o]||op&&cmp(make_pair(a0,a1),a[o]))
		return add(l,o-1,a0,a1,t,!op),void();
	add(o+1,r,a0,a1,t,!op);
}
int read()
{
	int s=0;
	char ch=getchar();
	while(ch<'0'||ch>'9')
		ch=getchar();
	while(ch>='0'&&ch<='9')
		s=s*10+ch-48,ch=getchar();
	return s;
}
void upd(int x)
{
	for(;x<=n;x+=x&-x)
		tr[x]++;
}
int query(int x)
{
	int ret=0;
	for(;x;x-=x&-x)
		ret+=tr[x];
	return ret;
}
void jia(int x,int op)
{
	ans+=op*abs(x-p[x]);
	int k=ask(1,m,x-1,p[x]);
	ans-=op*k;
	ans-=op*(p[x]-x+k);
}
int main()
{
	static int x[N],y[N],g[N];
	mnx[0]=mny[0]=1000000000;
	mxx[0]=mxy[0]=-1;
	n=read(),q=read();
	for(int i=1;i<=n;i++)
		g[i]=p[i]=read(),a[++m]=make_pair(i,p[i]);
	for(int i=1;i<=q;i++)
	{
		x[i]=read(),y[i]=read();
		swap(g[x[i]],g[y[i]]);
		a[++m]=make_pair(x[i],g[x[i]]);
		a[++m]=make_pair(y[i],g[y[i]]);
	}
	build(1,m,0);
	for(int i=1;i<=n;i++)
		add(1,m,i,p[i],1,0),upd(p[i]),ans+=abs(i-p[i])-i+query(p[i]);
	for(int i=1;i<=q;i++)
	{
		jia(x[i],-1);
		jia(y[i],-1);
		add(1,m,x[i],p[x[i]],-1,0);
		add(1,m,y[i],p[y[i]],-1,0);
		if(p[x[i]]>p[y[i]])
			--ans;
		swap(p[x[i]],p[y[i]]);
		add(1,m,x[i],p[x[i]],1,0);
		add(1,m,y[i],p[y[i]],1,0);
		jia(x[i],1);
		jia(y[i],1);
		if(p[x[i]]>p[y[i]])
			++ans;
		printf("%lld\n",ans);
	}
}