ZOJ - 4012(KDTree)

題意:給你n個座標點，m個查詢，每個查詢有直線與x軸，y軸的交點座標，然後讓你求這 條直線經過幾個座標點

題解：用kdtree對n個點進行切割，然後如果直線與能包圍該子樹點的最小矩形沒有交點那麼這條直線肯定沒有經過該子樹任何一點

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<stack>
using namespace std;
#define mes(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(i = a; i <= b; i++)
#define dec(i,a,b) for(i = b; i >= a; i--)
#define fi first
#define se second
#define ls rt<<1
#define rs rt<<1|1
#define mid (L+R)/2
#define lson ls,L,mid
#define rson rs,mid+1,R
typedef double db;
typedef long long int ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int mx = 1e5+5;
const int x_move[] = {1,-1,0,0,1,1,-1,-1};
const int y_move[] = {0,0,1,-1,1,-1,1,-1};
int n,m;
int flag;
struct node{
	int x,y;
	bool operator<(const node &a)const{
		if(flag) return x<a.x;
		return y < a.y;
	}
}a[mx];
struct kdtree{
	int l,r;
	int u,d;
	void operator=(const node &a){
		l = r = a.x;
		u = d = a.y;
	}
	kdtree operator+(const kdtree &a){
		kdtree k;
		k.l = min(a.l,l);
		k.r = max(a.r,r);
		k.u = max(a.u,u);
		k.d = min(a.d,d);
		return k;
	}
}kd[mx];
struct line{
	ll a,b,c;
	line(line &x){
		a = x.a;
		b = x.b;
		c = x.c;
	}
	line(ll x,ll y){
		a = y,b = x,c = a*b;
		c = -c;
	}
	bool inr(kdtree k){
		ll tmp = 1ll*a*k.l+c;
		tmp = -tmp;
		ll u = 1ll*k.u*b;
		ll d = 1ll*k.d*b;
		if(tmp>=d&&tmp<=u)
			return true;
		tmp = 1ll*a*k.r+c;
		tmp = -tmp;
		if(tmp>=d&&tmp<=u)
			return true;
		tmp = 1ll*b*k.u+c;
		tmp = -tmp;
		ll l = 1ll*k.l*a;
		ll r = 1ll*k.r*a;
		if(tmp>=l&&tmp<=r)
			return true;
		tmp = 1ll*b*k.d+c;
		tmp = -tmp;
		if(tmp>=l&&tmp<=r)
			return true;
		return false;
	}
	bool in(node t){
		//if(t.x==2){
		//	cout<<a<<b<<c<<endl;
	//		cout<<t.x*a+t.y*b+c<<endl;
//		}
		if(1ll*t.x*a+1ll*t.y*b+c==0)
			return true;
		return false;
	}
};
void built(int L,int R,int dep,int fa){
	if(L>R)
		return;
	flag = dep^1;
	nth_element(a+L,a+mid,a+R+1);
	kd[mid] = a[mid];
	built(L,mid-1,dep^1,mid);
	built(mid+1,R,dep^1,mid);
	kd[fa] = kd[mid]+kd[fa];
}
int query(int L,int R,line x){
	if(L>R)
		return 0;
	if(!x.inr(kd[mid]))
		return 0;
	int ans = 0;
	if(x.in(a[mid]))
		ans++;
	if(L==R)
		return ans;
	return query(L,mid-1,x)+ans+query(mid+1,R,x);
}
int main(){
	int t,q,ca = 1;
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&m);
		for(int i = 1; i <= n; i++)
			scanf("%d%d",&a[i].x,&a[i].y);
		built(1,n,1,0);
		int x,y;
		while(m--){
			scanf("%d%d",&x,&y);
			line l(x,y);
			printf("%d\n",query(1,n,l));
		}
	}
	return 0;
}