bzoj4520

KD-tree+堆

多年大坑

KD-tree已经是半年前学的了，忘记了。这道题当时一直T，今天重新抄了一遍，A了

KD-tree过程:1.建树:每次依次按x,y划分平面，像二叉搜索树一样建树，每个点维护一些东西；

2.查询:直接查太暴力了，我们用估价函数减值，每个点维护最小最大的x和y，每次计算能够造成的最大距离，如果有价值就递归。注意关于dl,dr的大小判断，这里query顺序不一样，因为先query大的会使大的先进堆，这样更快一些，相当于一个剪枝。其实KD-tree就是一个暴力剪枝的过程。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 1e18;
const int N = 1e5 + 5;
int n, k, root, d;
struct data {
    ll mx_x, mn_x, mx_y, mn_y, x, y, lc, rc;
    bool friend operator < (const data &a, const data &b) {
        if(d == 0) return a.x == b.x ? a.y < b.y : a.x < b.x;
        if(d == 1) return a.y == b.y ? a.x < b.x : a.y < b.y;
    }
} a[N];
priority_queue<ll, vector<ll>, greater<ll> > q;
ll sqr(ll x)
{
    return x * x;
}
ll dis(int k, ll x, ll y)
{
    return max(sqr(a[k].mn_x - x), sqr(a[k].mx_x - x)) + max(sqr(a[k].mn_y - y), sqr(a[k].mx_y - y));
}
void update(int x)
{
    a[x].mn_x = min(a[x].x, min(a[a[x].lc].mn_x, a[a[x].rc].mn_x));
    a[x].mx_x = max(a[x].x, max(a[a[x].lc].mx_x, a[a[x].rc].mx_x));
    a[x].mn_y = min(a[x].y, min(a[a[x].lc].mn_y, a[a[x].rc].mn_y));
    a[x].mx_y = max(a[x].y, max(a[a[x].lc].mx_y, a[a[x].rc].mx_y));
}
int build(int l, int r, int now)
{
    if(l > r) return 0;
    int mid = (l + r) >> 1;
    d = now;
    nth_element(a + l, a + mid, a + r + 1);
    a[mid].mx_x = a[mid].mn_x = a[mid].x;
    a[mid].mx_y = a[mid].mn_y = a[mid].y;
    a[mid].lc = build(l, mid - 1, now ^ 1);
    a[mid].rc = build(mid + 1, r, now ^ 1);
    update(mid);
    return mid;
}
void query(int k, ll x, ll y)
{
    ll tmp = sqr(x - a[k].x) + sqr(y - a[k].y), dl = a[k].lc ? dis(a[k].lc, x, y) : -inf, dr = a[k].rc ? dis(a[k].rc, x, y) : -inf;
    if(tmp > q.top()) q.pop(), q.push(tmp);
    if(dl < dr)
    {
        if(dr > q.top()) query(a[k].rc, x, y);
        if(dl > q.top()) query(a[k].lc, x, y);
    }
    else
    {
        if(dl > q.top()) query(a[k].lc, x, y);
        if(dr > q.top()) query(a[k].rc, x, y);
    }
}
int main()
{
    scanf("%d%d", &n, &k);
    a[0].mn_x = inf;
    a[0].mx_x = -inf;
    a[0].mn_y = inf;
    a[0].mx_y = -inf;
    for(int i = 1; i <= n; ++i) scanf("%d%d", &a[i].x, &a[i].y);
    root = build(1, n, 0);
    for(int i = 1; i <= 2 * k; ++i) q.push(0);
    for(int i = 1; i <= n; ++i) query(root, a[i].x, a[i].y);
    printf("%lld\n", q.top());
    return 0;
}