bzoj4520 Cqoi2016—K远点对

Posted 2020-08-27 MashiroSky

http://www.lydsy.com/JudgeOnline/problem.php?id=4520 (题目链接)

题意

　　求平面内第K远点对的距离。

Solution

　　左转题解：jump

细节

　　刚开始我还开了两个堆，想想其实是没必要的→_→

　　距离什么的开LL

代码

// bzoj4520
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<cmath>
#include<queue>
#define LL long long
#define inf 1<<30
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=100010;
int n,m,D,K,rt;
priority_queue<LL,vector<LL>,greater<LL> > q;

struct KDtree {
	int l,r,v[2],mn[2],mx[2];
	friend bool operator < (const KDtree a,const KDtree b) {
		return a.v[D]<b.v[D];
	}
}tr[maxn],S;

void update(int k) {
	for (int i=0;i<=1;i++) {
		if (tr[k].l) {
			tr[k].mn[i]=min(tr[k].mn[i],tr[tr[k].l].mn[i]);
			tr[k].mx[i]=max(tr[k].mx[i],tr[tr[k].l].mx[i]);
		}
		if (tr[k].r) {
			tr[k].mn[i]=min(tr[k].mn[i],tr[tr[k].r].mn[i]);
			tr[k].mx[i]=max(tr[k].mx[i],tr[tr[k].r].mx[i]);
		}
	}
}
int build(int l,int r,int p) {
	D=p;
	int mid=(l+r)>>1;
	nth_element(tr+l,tr+mid,tr+r+1);
	if (l<mid) tr[mid].l=build(l,mid-1,p^1);
	if (r>mid) tr[mid].r=build(mid+1,r,p^1);
	update(mid);
	return mid;
}
LL dis(LL x1,LL y1,LL x2,LL y2) {
	return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}
LL eva(int k) {
    LL lu=dis(S.v[0],S.v[1],tr[k].mn[0],tr[k].mn[1]);
    LL ru=dis(S.v[0],S.v[1],tr[k].mx[0],tr[k].mn[1]);
    LL ld=dis(S.v[0],S.v[1],tr[k].mn[0],tr[k].mx[1]);
    LL rd=dis(S.v[0],S.v[1],tr[k].mx[0],tr[k].mx[1]);
    return max(max(lu,ru),max(ld,rd));
}
void query(int k) {
	if (!k) return;
	LL d=dis(S.v[0],S.v[1],tr[k].v[0],tr[k].v[1]);
	if (q.top()<d) q.pop(),q.push(d);
	LL dl=eva(tr[k].l),dr=eva(tr[k].r);
	if (dl>dr) {
		if (q.top()<dl) query(tr[k].l);
		if (q.top()<dr) query(tr[k].r);
	}
	else {
		if (q.top()<dr) query(tr[k].r);
		if (q.top()<dl) query(tr[k].l);
	}
}
	
int main() {
	scanf("%d%d",&n,&K);
	for (int i=1;i<=n;i++) {
		scanf("%d%d",&tr[i].v[0],&tr[i].v[1]);
		tr[i].mn[0]=tr[i].mx[0]=tr[i].v[0];
		tr[i].mn[1]=tr[i].mx[1]=tr[i].v[1];
	}
	rt=build(1,n,0);
	for (int i=1;i<=2*K;i++) q.push(0);
	for (int i=1;i<=n;i++) S=tr[i],query(rt);
	printf("%lld",q.top());
	return 0;
}