[Cqoi2016]K远点对 K-Dtree

4520: [Cqoi2016]K远点对

链接

bzoj

思路

用K-Dtree求点的最远距离。
求的时候顺便维护一个大小为2k的小根堆。
不知道为啥一定会对。

代码

#include 
#define ll long long
#define ls (t[u].ch[0])
#define rs (t[u].ch[1])
#define cmin(a,b) (a>b?a=b:a)
#define cmax(a,b) (a>b?a:a=b)
using namespace std;
const int N=1e5+7;
const ll INF=1LL<<18;
int read() {
    int x=0,f=1;char s=getchar();
    for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;
    for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';
    return x*f;
}
priority_queue,greater > q;
int n,k,rt,WD;
struct point {
    int x[2];
    bool operator < (const point &b) const {
        return x[WD]r) return 0;
    int mid=(l+r)>>1;
    WD=wd;
    nth_element(p+l,p+mid,p+r+1);
    int u=mid;
    t[u].a=p[mid];
    ls=build(l,mid-1,wd^1);
    rs=build(mid+1,r,wd^1);
    up(u);
    return u;
}
inline ll fff(int a){return 1LL*a*a;}
inline ll dis(point a,point b) {
    return fff(a.x[0]-b.x[0])+fff(a.x[1]-b.x[1]);
}
ll get(int u,point a) {
    ll rec=0;
    for(int i=0;i<2;++i)
        rec+=max(fff(a.x[i]-t[u].ma[i]),fff(a.x[i]-t[u].mi[i]));
    return rec;
}
void query(point a,int u) {
    ll distance=dis(a,t[u].a);
    if(q.top()