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  • loj2043 「CQOI2016」K 远点对

    k-d tree 裸题………………

    #include <algorithm>
    #include <iostream>
    #include <cstdio>
    using namespace std;
    typedef long long ll;
    int n, k, rot, nowD, din;
    ll dui[205];
    const ll oo=9e18;
    struct Point{
    	int d[2], mn[2], mx[2], l, r;
    	int & operator[](int x){
    		return d[x];
    	}
    	bool operator<(const Point &x)const{
    		return d[nowD]<x.d[nowD];
    	}
    }p[100005], T;
    bool cmp(ll x, ll y){
    	return x>y;
    }
    struct KDTree{
    	Point t[100005];
    	void pushUp(int k){
    		int l=t[k].l, r=t[k].r;
    		for(int i=0; i<2; i++){
    			t[k].mn[i] = t[k].mx[i] = t[k][i];
    			if(l){
    				t[k].mn[i] = min(t[k].mn[i], t[l].mn[i]);
    				t[k].mx[i] = max(t[k].mx[i], t[l].mx[i]);
    			}
    			if(r){
    				t[k].mn[i] = min(t[k].mn[i], t[r].mn[i]);
    				t[k].mx[i] = max(t[k].mx[i], t[r].mx[i]);
    			}
    		}
    	}
    	int build(int l, int r, int now){
    		nowD = now;
    		int mid=(l+r)>>1;
    		nth_element(p+l, p+mid, p+r+1);
    		t[mid] = p[mid];
    		if(l<mid)	t[mid].l = build(l, mid-1, now^1);
    		if(mid<r)	t[mid].r = build(mid+1, r, now^1);
    		pushUp(mid);
    		return mid;
    	}
    	ll getDis(const Point &u, const Point &v){
    		return (ll)(u.d[1]-v.d[1])*(u.d[1]-v.d[1])+(ll)(u.d[0]-v.d[0])*(u.d[0]-v.d[0]);
    	}
    	ll simDis(int k){
    		ll re=0;
    		for(int i=0; i<2; i++){
    			ll tmp=max(abs(t[k].mn[i]-T[i]), abs(t[k].mx[i]-T[i]));
    			re += tmp * tmp;
    		}
    		return re;
    	}
    	void query(int k){
    		ll d=getDis(t[k], T), disl=-oo, disr=-oo;
    		int l=t[k].l, r=t[k].r;
    		if(d>dui[1]){
    			pop_heap(dui+1, dui+1+din, cmp);
    			dui[din] = d;
    			push_heap(dui+1, dui+1+din, cmp);
    		}
    		if(l)	disl = simDis(l);
    		if(r)	disr = simDis(r);
    		if(disl>disr){
    			if(disl>dui[1])	query(l);
    			if(disr>dui[1])	query(r);
    		}
    		else{
    			if(disr>dui[1])	query(r);
    			if(disl>dui[1])	query(l);	
    		}
    	}
    }kdt;
    int main(){
    	cin>>n>>k;
    	k *= 2;
    	for(int i=1; i<=n; i++)
    		scanf("%d %d", &p[i][0], &p[i][1]);
    	rot = kdt.build(1, n, 0);
    	while(k--)	dui[++din] = -oo;
    	for(int i=1; i<=n; i++){
    		T = p[i];
    		kdt.query(rot);
    	}
    	cout<<dui[1]<<endl;
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/poorpool/p/8952624.html
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