P3157 [CQOI2011]动态逆序对

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#pragma GCC optimize(2)
#define up(i,a,b)  for(int i=a;i<b;i++)
#define dw(i,a,b)  for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
	char ch = getchar(); ll x = 0, f = 1;
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
int a[N], b[N], vis[N], rk[N];
ll ans[N];
struct BIT {
	ll sum[N];
	int lim;
	int lowbit(int i)
	{
		return i & (-i);
	}
	void update(int pos,ll val)
	{
		while (pos <= lim)
		{
			sum[pos] += val;
			pos += lowbit(pos);
		}
	}
	ll query(int pos)
	{
		ll res = 0;
		while (pos)
		{
			res += sum[pos];
			pos -= lowbit(pos);
		}
		return res;
	}
}T;
struct node
{
	int x, y, z;
}per[N];
int num;
bool cmpx(node a, node b)
{
	return a.x < b.x;
}
bool cmpy(node a, node b)
{
	return a.y < b.y;
}
int n, m;
void cdq(int l, int r)
{	
	if (l == r)return;
	int mid = (l + r) >> 1;
	cdq(l, mid);
	cdq(mid + 1, r);
	sort(per + l, per + mid + 1, cmpy);
	sort(per + mid + 1, per + r + 1, cmpy);
	for (int i = r, j = mid; i >= mid+1; i--)
	{
		while (j>=l&&per[j].y > per[i].y) {
			T.update(per[j].z, 1);
			j--;
		}
		ans[per[i].x] += T.query(per[i].z);
	}
	for (int i = r, j = mid; i >= mid+1;i--)
	{
		while (j >=l && per[j].y > per[i].y) {
			T.update(per[j].z, -1);
			j--;
		}
	}
	for (int i = mid + 1, j = l; i <= r; i++)
	{
		while (j <= mid && per[j].y < per[i].y) {
			T.update(per[j].z, 1);
			j++;
		}
		ans[per[i].x] += (T.query(n) - T.query(per[i].z));
	}
	for (int i = mid + 1, j = l; i <= r; i++)
	{
		while (j <= mid && per[j].y < per[i].y) {
			T.update(per[j].z, -1);
			j++;
		}
	}
}
int main()
{
	n = read(), m = read();
	int u;
	upd(i, 1, n)
	{
		u = read();
		a[i] = u;
		rk[u] = i;
	}
	upd(i, 1, m)
	{
		u = read();
		b[i] = u; vis[u] = 1;
	}
	upd(i, 1, n)
	{
		if (!vis[a[i]])
		{
			per[++num].x = num;
			per[num].y = i;
			per[num].z = a[i];
		}
	}
	dwd(i, m, 1)
	{
		per[++num].x = num;
		per[num].y = rk[b[i]];
		per[num].z = b[i];
	}
	sort(per + 1, per + num + 1, cmpx);
	T.lim = n;
	cdq(1, n);
	upd(i, 1, n)
	{
		ans[i] += ans[i - 1];
		//cout <<i<<" "<< ans[i] << endl;
	}
	int j = n;
	dwd(i, m, 1)
	{
		printf("%lld\n", ans[j--]);
	}
	return 0;
}