[USACO17FEB]Why Did the Cow Cross the Road III P

[USACO17FEB]Why Did the Cow Cross the Road III P

考虑我们对每种颜色记录这样一个信息 \((x,y,z)\)，即左边出现的位置，右边出现的位置，该颜色。

于是统计的是\(x < x_2,y > y_2,|z - z2| > k\)的数对数量。

因为\(CDQ\)分治的过程是，第一维事先排序，第二维递归进行，第三维用数据结构统计，所以

上述这个数量是很好处理的。

// Problem: P3658 [USACO17FEB]Why Did the Cow Cross the Road III P
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3658
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
#define ll long long
#define N 1000010

ll n,k,ans;

ll to[N];

struct P{
	int x,y,z;
}a[N],b[N];

bool operator < (P aa,P bb){
	return aa.x == bb.x ? ((aa.y == bb.y) ? (aa.z < bb.z) : aa.y > bb.y) : aa.x < bb.x;
}


ll t[N];

#define lowbit(x) (x & -x)

inline void add(int x,int u){
	for(int i = x;i <= n;i += lowbit(i))
	t[i] += u;
}

inline ll q(int x){
	x = std::max(x,0);
	x = std::min(n,(ll)x);
	ll ans = 0;
	for(int i = x;i;i -= lowbit(i))
	ans += t[i];
	return ans;
}

//_________BIT

#define mid ((l + r) >> 1)

inline void change(int l,int r){
	if(l == r)return;
	int i = l,j = mid + 1,p = l - 1;
	while(i <= mid && j <= r){if(a[i].y > a[j].y)b[++p] = a[i++];else b[++p] = a[j++];}
	while(i <= mid)b[++p] = a[i++];
	while(j <= r)b[++p] = a[j++];
	for(int k = l;k <= r;++k)
	a[k] = b[k];
}

inline void solve(int l,int r){
	if(l == r)return ;
	solve(l,mid);solve(mid + 1,r);change(l,mid);change(mid + 1,r);
	int i = l,j = mid + 1;
	while(j <= r){
		while(i <= mid && a[i].y > a[j].y)add(a[i++].z,1);
		ans = ans + (ll)q(a[j].z - k - 1) + (ll)q(n) - (ll)q(a[j].z + k);
		++j;
	}
	for(int k = l;k < i;++k)
	add(a[k].z,-1);
}

//————————————————————cdq
int main(){
	scanf("%lld%lld",&n,&k);
	for(int i = 1;i <= n;++i){
		ll x;
		scanf("%lld",&x);
		to[x] = i;
	}
	for(int i = 1;i <= n;++i){
		ll x;
		scanf("%lld",&x);
		a[i].x = to[x],a[i].y = i,a[i].z = x;
	}
	std::sort(a + 1,a + n + 1);
	solve(1,n);
	std::cout<<ans<<std::endl;
	return 0;
}