Bzoj4520: [Cqoi2016]K远点对

题面

Bzoj

Sol

维护一个小根堆，初始里面放 2∗k 2 ∗ k 个元素(因为点对可能算两遍)
每个点 KDTree K D T r e e 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 
# 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  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;
}