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  • Bzoj4520: [Cqoi2016]K远点对

    题面

    Bzoj

    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
    ", -q.top());
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/9113805.html
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