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\n",q.top().id);
    }
    getchar();getchar();
    return 0;
}

转载于:https://www.cnblogs.com/Skyminer/p/6427217.html