bzoj 2626: JZPFAR k-D树

题目大意:

平面上n个点,每次给出一个点,求这个点的k远点

题解:

什么叫做k远点呢。。。
1 2 3 4 5中5是第一远,4是第二远...
看来我语文学的不好
那么我们直接上k-D Tree求k邻近的方式求k远离即可

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline void read(ll &x){
	x=0;char ch;bool flag = false;
	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline ll max(ll a,ll b,ll c){return max(a,max(b,c));}
inline ll min(ll a,ll b,ll c){return min(a,min(b,c));}
const ll maxn = 100010;
const ll dem = 2;
struct Node{
	Node *ch[2];
	ll pos[2];ll id;
	ll minn[2],maxx[2];
	void update(){
		for(ll d=0;d<dem;++d){
			minn[d] = min(pos[d],ch[0]->minn[d],ch[1]->minn[d]);
			maxx[d] = max(pos[d],ch[0]->maxx[d],ch[1]->maxx[d]);
		}
	}
}*null;
Node akdjalskjalksdlkasdlkjlkjflsgdjasdf;
Node T[maxn];
inline void init(){
	null = &akdjalskjalksdlkasdlkjlkjflsgdjasdf;
	null->ch[0] = null->ch[1] = null;
	for(ll d=0;d<dem;++d){
		null->pos[d] = 0;
		null->minn[d] = 0x3f3f3f3f;
		null->maxx[d] = -0x3f3f3f3f;
	}
}
ll now,split[maxn];
inline bool cmp(const Node &a,const Node &b){
	return a.pos[split[now]] < b.pos[split[now]];
}
Node* build(ll l,ll r,ll s){
	if(l > r) return null;
	ll mid = (l+r) >> 1;
	split[now = mid] = s % dem;
	nth_element(T+l,T+mid,T+r+1,cmp);
	Node *p = &T[mid];
	p->ch[0] = build(l,mid-1,s+1);
	p->ch[1] = build(mid+1,r,s+1);
	p->update();return p;
}
struct Data{
	ll dis;ll id;
	bool operator < (const Data &a)const{
		if(dis != a.dis) return dis > a.dis;
		return id < a.id;
	}
	Data(){}
	Data(ll a,ll b){dis=a;id=b;}
};
priority_queue<Data>q;
inline ll sqr(ll x){return x*x;}
Node op;ll k;
inline ll md(Node *p){
	ll ret = .0;
	ret = max(ret,sqr(p->minn[0] - op.pos[0]) + sqr(p->minn[1] - op.pos[1]));
	ret = max(ret,sqr(p->minn[0] - op.pos[0]) + sqr(p->maxx[1] - op.pos[1]));
	ret = max(ret,sqr(p->maxx[0] - op.pos[0]) + sqr(p->minn[1] - op.pos[1]));
	ret = max(ret,sqr(p->maxx[0] - op.pos[0]) + sqr(p->maxx[1] - op.pos[1]));
	return ret;
}
void query(Node *p){
	if(p == null) return;
	ll dis = 0;
	for(ll d=0;d<dem;++d) dis += sqr(op.pos[d] - p->pos[d]);
	if(q.size() < k){
		q.push(Data(dis,p->id));
	}else if(Data(dis,p->id) < q.top()){
		q.pop();q.push(Data(dis,p->id));
	}
	if(md(p->ch[0]) < md(p->ch[1])) swap(p->ch[0],p->ch[1]);
	
	if(p->ch[0] != null && ((q.size() < k) || (md(p->ch[0]) >= q.top().dis))) query(p->ch[0]);
	if(p->ch[1] != null && ((q.size() < k) || (md(p->ch[1]) >= q.top().dis))) query(p->ch[1]);
}
int main(){
	init();
	ll n;read(n);
	for(ll i=1;i<=n;++i){
		for(ll d=0;d<dem;++d){
			read(T[i].pos[d]);
		}
		T[i].ch[0] = T[i].ch[1] = null;
		T[i].id = i;T[i].update();
	}
	Node *root = build(1,n,1);
	ll m;read(m);
	while(m--){
		for(ll d=0;d<dem;++d) read(op.pos[d]);
		read(k);
		while(!q.empty()) q.pop();
		query(root);
		printf("%lld
",q.top().id);
	}
	getchar();getchar();
	return 0;
}