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; }
原文:http://www.cnblogs.com/19992147orz/p/7868625.html