嘟嘟嘟
做过[国家集训队]JZPFAR这道题的话，这题就不难了。
我们维护一个长度为\(k\)的小根堆，在加入第\(i\)个点之前，用\([1, i - 1]\)这些点离点\(i\)的距离更新答案。这样也能保证每一对点之间的距离一定只算了一次。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e5 + 5;
const int SIZE = 5000;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ' ';
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
if(last == '-') ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar('-');
if(x >= 10) write(x / 10);
putchar(x % 10 + '0');
}
int n, K, Dim;
priority_queue<ll, vector<ll>, greater<ll> > q;
struct Tree
{
int ch[2];
ll d[2], Min[2], Max[2];
In bool operator < (const Tree& oth)const
{
return d[Dim] < oth.d[Dim];
}
}t[maxn], a[maxn];
int root, tcnt = 0, cnt = 0;
In void pushup(int now)
{
for(int i = 0; i < 2; ++i)
{
if(t[now].ch[0])
{
t[now].Min[i] = min(t[now].Min[i], t[t[now].ch[0]].Min[i]);
t[now].Max[i] = max(t[now].Max[i], t[t[now].ch[0]].Max[i]);
}
if(t[now].ch[1])
{
t[now].Min[i] = min(t[now].Min[i], t[t[now].ch[1]].Min[i]);
t[now].Max[i] = max(t[now].Max[i], t[t[now].ch[1]].Max[i]);
}
}
}
In void new_node(int& now, Tree a)
{
t[now = ++tcnt] = a;
for(int i = 0; i < 2; ++i)
{
t[now].ch[i] = 0;
t[now].Min[i] = t[now].Max[i] = t[now].d[i];
}
}
In void build(int& now, int L, int R, int d)
{
if(L > R) return;
int mid = (L + R) >> 1;
Dim = d;
nth_element(a + L, a + mid, a + R + 1);
new_node(now, a[mid]);
build(t[now].ch[0], L, mid - 1, d ^ 1);
build(t[now].ch[1], mid + 1, R, d ^ 1);
pushup(now);
}
In void insert(int& now, Tree a, int d)
{
if(!now) {new_node(now, a); return;}
if(a.d[d] <= t[now].d[d]) insert(t[now].ch[0], a, d ^ 1);
else insert(t[now].ch[1], a, d ^ 1);
pushup(now);
}
In ll dis(int now, ll * d)
{
ll ret = 0;
for(int i = 0; i < 2; ++i) ret += (d[i] - t[now].d[i]) * (d[i] - t[now].d[i]);
return ret;
}
In ll price(int now, ll* d)
{
ll ret = 0;
for(int i = 0; i < 2; ++i)
{
ll Max = max(abs(t[now].Min[i] - d[i]), abs(t[now].Max[i] - d[i]));
ret += Max * Max;
}
return ret;
}
In void query(int now, ll* d)
{
if(!now) return;
ll tp = dis(now, d);
if(tp > q.top()) q.pop(), q.push(tp);
ll disL = price(t[now].ch[0], d), disR = price(t[now].ch[1], d);
if(disL < disR) swap(t[now].ch[0], t[now].ch[1]), swap(disL, disR);
if(disL > q.top()) query(t[now].ch[0], d);
if(disR > q.top()) query(t[now].ch[1], d);
}
int main()
{
n = read(); K = read();
for(int i = 1; i <= K; ++i) q.push(-1);
for(int i = 1; i <= n; ++i)
{
a[++cnt].d[0] = read(), a[cnt].d[1] = read();
query(root, a[cnt].d);
if(cnt % SIZE == 0)
{
for(int i = 1; i <= tcnt; ++i) t[i].ch[0] = t[i].ch[1] = 0;
tcnt = 0;
build(root, 1, cnt, 0);
}
else insert(root, a[cnt], 0);
}
write(q.top()), enter;
return 0;
}