bzoj4520 K远点对

题目链接

思路

这个"\(K\)远“点对一直理解成了距离第\(K\)大的点对\(233\)

要求第\(K\)远,那么我们只要想办法求出来最远的\(K\)个点对就可以了。

用一个大小为\(2K\)(因为每个点对会被统计两次)的小头堆维护距离最大的\(K\)个点对,然后在\(KD-tree\)上查询最远点对,如果查到的点对之间的距离比堆顶大,那么就把堆顶弹出来,当前距离插进去。

最后堆顶元素就是答案。

代码

/*
* @Author: wxyww
* @Date:   2019-06-13 07:45:59
* @Last Modified time: 2019-06-13 08:47:21
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
#define int ll
const int N = 100100,INF = 1e9;
#define ls TR[rt].ch[0]
#define rs TR[rt].ch[1]
priority_queue<int,vector<int>,greater<int> >q;
ll read() {
	ll x=0,f=1;char c=getchar();
	while(c<'0'||c>'9') {
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9') {
		x=x*10+c-'0';
		c=getchar();
	}
	return x*f;
}
int mul(int x) {
	return x * x;
}
struct node {
	int ch[2],d[2],mx[2],mn[2];
}TMP,TR[N],a[N];
int D;
bool cmp(const node &A,const node &B) {
	return A.d[D] < B.d[D];
}
void up(int rt) {
	for(int i = 0;i <= 1;++i) {
		if(ls) TR[rt].mx[i] = max(TR[ls].mx[i],TR[rt].mx[i]),
			TR[rt].mn[i] = min(TR[ls].mn[i],TR[rt].mn[i]);
		if(rs) TR[rt].mx[i] = max(TR[rs].mx[i],TR[rt].mx[i]),
			TR[rt].mn[i] = min(TR[rs].mn[i],TR[rt].mn[i]);
	}
}
int build(int l,int r,int now) {
	D = now;
	int mid = (l + r) >> 1;
	nth_element(a + l,a + mid,a + r + 1,cmp);
	TR[mid] = a[mid];
	for(int i = 0;i <= 1;++i) TR[mid].mx[i] = TR[mid].mn[i] = TR[mid].d[i];
	int rt = mid;
	if(l < mid) ls = build(l,mid - 1,now ^ 1);
	if(r > mid) rs = build(mid + 1,r,now ^ 1);
	up(mid);
	return mid;
}
int dis(const node &A,const node &B) {
	return mul(A.d[0] - B.d[0]) + mul(A.d[1] - B.d[1]);
}
int get(const node &A,const node &B) {
	int ret = 0;
	for(int i = 0;i <= 1;++i)
		ret += mul(max(abs(A.d[i] - B.mx[i]),abs(A.d[i] - B.mn[i])));
	return ret;
}
void query(int rt) {
	int K = dis(TMP,TR[rt]);
	if(K > q.top()) q.pop(),q.push(K);
	int dl = -INF,dr = -INF;
	if(ls) dl = get(TMP,TR[ls]);
	if(rs) dr = get(TMP,TR[rs]);
	if(dl > dr) {
		if(dl > q.top()) query(ls);
		if(dr > q.top()) query(rs);
	}
	if(dr > dl) {
		if(dr > q.top()) query(rs);
		if(dl > q.top()) query(ls);
	}
}
int X[N],Y[N];
signed main() {
	int n = read(),K = read();
	for(int i = 1;i <= n;++i) {
		X[i] = a[i].d[0] = read();Y[i] = a[i].d[1] = read();
	}
	int root = build(1,n,0);
	for(int i = 1;i <= K * 2;++i) q.push(0);
	for(int i = 1;i <= n;++i) {
		TMP.d[0] = X[i],TMP.d[1] = Y[i];
		query(root);
	}
	cout<<q.top();
	return 0;
}
posted @ 2019-06-13 08:50  wxyww  阅读(...)  评论(...编辑  收藏