P3157 [CQOI2011]动态逆序对

P3157 [CQOI2011]动态逆序对
原题地址
此处写的是cdq分治板子题（三维偏序）

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define For(i,x,y)	for(ll i=(x);i<=(y);++i)
#define FOr(i,x,y)	for(ll i=(x);i>=(y);--i)
#define rep(i,x,y)	for(ll i=(x);i<(y);++i)
#define clr(a,v)	memset(a,v,sizeof(a))
#define cpy(a,b)	memcpy(a,b,sizeof(a))
#define fi	first
#define se	second
#define	pb	push_back
#define mk	make_pair
#define pa	pair<ll,ll>
#define y1	y11111111111111
#define debug	puts("@@@@@@@@@@@@@@@@@@@@@@@")
ll read(){
	ll x=0,f=1;	char ch=getchar();
	for(;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for(;ch>='0'&&ch<='9';ch=getchar())	x=x*10+ch-'0';
	return x*f;
}
void write(ll x){
	if (x<0)	putchar('-'),write(-x);
	else{
		if (x>=10)	write(x/10);
		putchar(x%10+'0');
	}
}
#define lowbit(x) (x&(-x))
const int N = 200005;
ll n,m;
ll b[N];
struct node{
	ll val,ans,del;
}a[N];
ll res[N];

inline void change(ll x,ll v){ for(;x<=n+1;x += lowbit(x)) b[x] += v; }
inline ll query(ll x){ll ans = 0; for(;x;x -= lowbit(x)) ans += b[x]; return ans;}

inline bool cmp(node a,node b){
	return a.del < b.del;
}
inline bool cmp2(node a,node b){
	return a.val < b.val;
}

inline void cdq(ll l,ll r){
	if(l == r) return;
	ll mid = (l+r)>>1;
	cdq(l,mid); cdq(mid+1,r);
	ll i = l,j = mid+1;
	while(i <= mid){
		while(a[i].val > a[j].val && j <= r) change(a[j].del,1),++j;
		a[i].ans += query(m) - query(a[i].del-1);
		++i;
	}
	i = l,j = mid+1;
	while(i <= mid){
		while(a[i].val > a[j].val && j <= r) change(a[j].del,-1),++j;
		++i;
	}
	i = mid; j = r;
	while(j >= mid+1){
		while(a[j].val < a[i].val && i >= l) change(a[i].del,1),--i;
		a[j].ans += query(m) - query(a[j].del-1);
		--j;
	}
	i = mid; j = r;
	while(j >= mid+1){
		while(a[j].val < a[i].val && i >= l) change(a[i].del,-1),--i;
		--j;
	}
	sort(a+l,a+r+1,cmp2);		//偷懒使用sort
}

signed main(){
	n = read(),m = read();
	For(i,1,n){
		a[i].val = read();
		res[a[i].val] = i;
	}
	For(i,1,m){
		ll x = read();
		a[res[x]].del = i;
	}
	For(i,1,n) if(a[i].del == 0) a[i].del = m+1;
	ll ans = 0;
	For(i,1,n){
		ans += query(n)-query(a[i].val-1);
		change(a[i].val,1);
	}
	For(i,1,n) change(a[i].val,-1);
	m = m+1;
	cdq(1,n);
	sort(a+1,a+n+1,cmp);
	rep(i,1,m){
		printf("%lld\n",ans);
		ans -= a[i].ans;
	}
}