题面
Sol
维护一个小根堆，初始里面放\(2*k\)个元素(因为点对可能算两遍)
每个点\(KDTree\)暴力查询是否有与这个点距离大于堆顶的，替换堆顶就好了
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
const int maxn(1e5 + 5);
int n, k, op, rt, num, cnt;
priority_queue <ll> q;
struct Point{
int d[2];
IL int operator <(RG Point a) const{
return d[op] < a.d[op];
}
} a[maxn];
struct KDTree{
int ch[2], d[2], mn[2], mx[2];
} tr[maxn];
IL void Chkmax(RG int &x, RG int y){
if(y > x) x = y;
}
IL void Chkmin(RG int &x, RG int y){
if(y < x) x = y;
}
IL void Update(RG int x){
RG int ls = tr[x].ch[0], rs = tr[x].ch[1];
tr[x].mx[0] = tr[x].mn[0] = tr[x].d[0];
tr[x].mx[1] = tr[x].mn[1] = tr[x].d[1];
if(ls){
Chkmin(tr[x].mn[0], tr[ls].mn[0]), Chkmin(tr[x].mn[1], tr[ls].mn[1]);
Chkmax(tr[x].mx[0], tr[ls].mx[0]), Chkmax(tr[x].mx[1], tr[ls].mx[1]);
}
if(rs){
Chkmin(tr[x].mn[0], tr[rs].mn[0]), Chkmin(tr[x].mn[1], tr[rs].mn[1]);
Chkmax(tr[x].mx[0], tr[rs].mx[0]), Chkmax(tr[x].mx[1], tr[rs].mx[1]);
}
}
IL int Build(RG int l, RG int r, RG int nop){
op = nop;
RG int x = (l + r) >> 1, nw = ++cnt;
nth_element(a + l, a + x, a + r + 1);
tr[nw].d[0] = a[x].d[0], tr[nw].d[1] = a[x].d[1];
if(l < x) tr[nw].ch[0] = Build(l, x - 1, nop ^ 1);
if(x < r) tr[nw].ch[1] = Build(x + 1, r, nop ^ 1);
Update(nw);
return nw;
}
# define Sqr(x) ((x) * (x))
IL ll Dis(RG ll x1, RG ll y1, RG ll x2, RG ll y2){
return Sqr(x1 - x2) + Sqr(y1 - y2);
}
IL ll Calc(RG int p, RG ll x, RG ll y){
RG ll ret = 0;
ret = max(ret, Dis(tr[p].mn[0], tr[p].mn[1], x, y));
ret = max(ret, Dis(tr[p].mx[0], tr[p].mn[1], x, y));
ret = max(ret, Dis(tr[p].mn[0], tr[p].mx[1], x, y));
ret = max(ret, Dis(tr[p].mx[0], tr[p].mx[1], x, y));
return ret;
}
IL void Query(RG int x, RG Point p){
RG ll dis = Dis(tr[x].d[0], tr[x].d[1], p.d[0], p.d[1]);
if(dis > -q.top()) q.pop(), q.push(-dis);
if(tr[x].ch[0]){
if(Calc(tr[x].ch[0], p.d[0], p.d[1]) > -q.top()) Query(tr[x].ch[0], p);
}
if(tr[x].ch[1]){
if(Calc(tr[x].ch[1], p.d[0], p.d[1]) > -q.top()) Query(tr[x].ch[1], p);
}
}
int main(){
n = Input(), k = Input();
for(RG int i = 1; i <= n; ++i) a[i].d[0] = Input(), a[i].d[1] = Input();
rt = Build(1, n, 0);
for(RG int i = k + k; i; --i) q.push(0);
for(RG int i = 1; i <= n; ++i) Query(rt, a[i]);
printf("%lld\n", -q.top());
return 0;
}