BZOJ 4520: [Cqoi2016]K远点对

KD-Tree暴力大法吼哇！

容易发现如果我们给每个点求出\(K\)个最远距离，放在一起找出其中的第\(2K\)大的就是答案（一对点会算两次）

考虑搞出一个小根堆，刚开始往里面放\(2K\)个\(0\)。然后我们枚举每个点，不断地找出距离它的最远点然后和堆顶比较，如果大于就替换掉堆顶

最后答案就是堆顶，而这种做法正确性显然且有一定剪枝，但复杂度是\(O(n^2)\)级别的

然后我们就KD-Tree来实现暴力的过程即可（滑稽），复杂度\(O(\text{玄学})\)，实测最慢跑了70ms。。。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#define RI register int
#define CI const int&
using namespace std;
typedef long long LL;
const int N=100005; const LL INF=1e18;
int D; priority_queue < LL,vector <LL>,greater <LL> > hp;
struct point
{
    int d[2];
    friend inline bool operator < (const point& A,const point& B)
    {
        return A.d[D]==B.d[D]?A.d[D^1]<B.d[D^1]:A.d[D]<B.d[D];
    }
}a[N],s; int n,k,rt;
inline LL dist(const point& A,const point& B)
{
    return 1LL*(A.d[0]-B.d[0])*(A.d[0]-B.d[0])+1LL*(A.d[1]-B.d[1])*(A.d[1]-B.d[1]);
}
class KD_Tree
{
    private:
        struct kd_interval
        {
            int ch[2]; point p,mi,mx;
        }node[N]; int tot;
        #define lc(x) node[x].ch[0]
        #define rc(x) node[x].ch[1]
        #define P(x) node[x].p
        #define Mi(x) node[x].mi
        #define Mx(x) node[x].mx
        inline void pushup(CI x,CI y)
        {
            for (RI i=0;i<2;++i) Mi(x).d[i]=min(Mi(x).d[i],Mi(y).d[i]),
            Mx(x).d[i]=max(Mx(x).d[i],Mx(y).d[i]);
        }
        inline LL getmax(CI now)
        {
            if (!now) return -INF; LL cur=-INF;
            cur=max(cur,dist((point){Mi(now).d[0],Mi(now).d[1]},s));
            cur=max(cur,dist((point){Mi(now).d[0],Mx(now).d[1]},s));
            cur=max(cur,dist((point){Mx(now).d[0],Mi(now).d[1]},s));
            cur=max(cur,dist((point){Mx(now).d[0],Mx(now).d[1]},s)); return cur;
        }
    public:
        inline void build(int& now,CI l=1,CI r=n,CI d=0)
        {
            now=++tot; int mid=l+r>>1; D=d; nth_element(a+l+1,a+mid+1,a+r+1);
            P(now)=Mi(now)=Mx(now)=a[mid];
            if (l!=mid) build(lc(now),l,mid-1,d^1),pushup(now,lc(now));
            if (r!=mid) build(rc(now),mid+1,r,d^1),pushup(now,rc(now));
        }
        inline void querymax(CI now)
        {
            if (!now) return; if (getmax(now)<hp.top()) return;
            LL tp=dist(s,P(now)); if (tp>hp.top()) hp.pop(),hp.push(tp);
            LL ml=getmax(lc(now)),mr=getmax(rc(now));
            if (ml>mr) querymax(lc(now)),querymax(rc(now));
            else querymax(rc(now)),querymax(lc(now));
        }
        #undef lc
        #undef rc
        #undef P
        #undef Mi
        #undef Mx
}KD;
int main()
{
    RI i; for (scanf("%d%d",&n,&k),i=1;i<=n;++i) scanf("%d%d",&a[i].d[0],&a[i].d[1]);
    for (KD.build(rt),i=1;i<=2*k;++i) hp.push(0); for (i=1;i<=n;++i)
    s=(point){a[i].d[0],a[i].d[1]},KD.querymax(rt); return printf("%lld",hp.top()),0;
}

BZOJ 4520: [Cqoi2016]K远点对

原文：https://www.cnblogs.com/cjjsb/p/12250474.html