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kd_tree.go
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kd_tree.go
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package copypasta
import "math/rand"
/* k-d tree: k-dimensional tree; k 维树
https://en.wikipedia.org/wiki/K-d_tree
推荐 https://www.luogu.com.cn/blog/command-block/kdt-xiao-ji
https://www.luogu.com.cn/blog/lc-2018-Canton/solution-p4148
https://oi-wiki.org/ds/kdt/
todo 题单 https://www.luogu.com.cn/training/4295
模板题 https://www.luogu.com.cn/problem/P4148
todo https://codeforces.com/problemset/problem/44/G
*/
type kdNode struct {
lr [2]*kdNode
p, mi, mx [2]int // 0 为 x,1 为 y
sz, val, sm int
}
func (o *kdNode) size() int {
if o != nil {
return o.sz
}
return 0
}
func (o *kdNode) sum() int {
if o != nil {
return o.sm
}
return 0
}
func (o *kdNode) maintain() {
o.sz = o.lr[0].size() + o.lr[1].size() + 1
o.sm = o.lr[0].sum() + o.lr[1].sum() + o.val
for i := 0; i < 2; i++ {
o.mi[i] = o.p[i]
o.mx[i] = o.p[i]
for _, ch := range o.lr {
if ch != nil {
o.mi[i] = min(o.mi[i], ch.mi[i])
o.mx[i] = max(o.mx[i], ch.mx[i])
}
}
}
}
func (o *kdNode) nodes() []*kdNode {
nodes := make([]*kdNode, 0, o.size())
var f func(*kdNode)
f = func(o *kdNode) {
if o != nil {
nodes = append(nodes, o)
f(o.lr[0])
f(o.lr[1])
}
}
f(o)
rand.Shuffle(len(nodes), func(i, j int) { nodes[i], nodes[j] = nodes[j], nodes[i] })
return nodes
}
func divideKDT(a []*kdNode, k, dim int) {
for l, r := 0, len(a)-1; l < r; {
v := a[l].p[dim]
i, j := l, r+1
for {
for i++; i < r && a[i].p[dim] < v; i++ {
}
for j--; j > l && a[j].p[dim] > v; j-- {
}
if i >= j {
break
}
a[i], a[j] = a[j], a[i]
}
a[l], a[j] = a[j], a[l]
if j == k {
break
} else if j < k {
l = j + 1
} else {
r = j - 1
}
}
}
// 另一种实现是选择的维度要满足其内部点的分布的差异度最大,见 https://oi-wiki.org/ds/kdt/
func buildKDT(nodes []*kdNode, dim int) *kdNode {
if len(nodes) == 0 {
return nil
}
m := len(nodes) / 2
divideKDT(nodes, m, dim)
o := nodes[m]
o.lr[0] = buildKDT(nodes[:m], dim^1)
o.lr[1] = buildKDT(nodes[m+1:], dim^1)
o.maintain()
return o
}
func (o *kdNode) rebuild(dim int) *kdNode { return buildKDT(o.nodes(), dim) }
func (o *kdNode) put(p [2]int, val, dim int) *kdNode {
if o == nil {
o = &kdNode{p: p, val: val}
o.maintain()
return o
}
if p[dim] < o.p[dim] {
o.lr[0] = o.lr[0].put(p, val, dim^1)
} else {
o.lr[1] = o.lr[1].put(p, val, dim^1)
}
o.maintain()
if sz := o.size() * 3; o.lr[0].size()*4 > sz || o.lr[1].size()*4 > sz { // alpha=3/4
return o.rebuild(dim)
}
return o
}
// 矩形 X-Y 在矩形 x-y 内
func inRect(x1, y1, x2, y2, X1, Y1, X2, Y2 int) bool {
return x1 <= X1 && X2 <= x2 && y1 <= Y1 && Y2 <= y2
}
// 矩形 X-Y 在矩形 x-y 外
func outRect(x1, y1, x2, y2, X1, Y1, X2, Y2 int) bool {
return X2 < x1 || X1 > x2 || Y2 < y1 || Y1 > y2
}
func (o *kdNode) query(x1, y1, x2, y2 int) (res int) {
if o == nil || outRect(x1, y1, x2, y2, o.mi[0], o.mi[1], o.mx[0], o.mx[1]) {
return
}
if inRect(x1, y1, x2, y2, o.mi[0], o.mi[1], o.mx[0], o.mx[1]) {
return o.sm
}
if inRect(x1, y1, x2, y2, o.p[0], o.p[1], o.p[0], o.p[1]) { // 根在询问矩形内
res = o.val
}
res += o.lr[0].query(x1, y1, x2, y2) + o.lr[1].query(x1, y1, x2, y2)
return
}
type kdTree struct {
root *kdNode
}
func newKdTree() *kdTree {
return &kdTree{}
}
func (t *kdTree) put(p [2]int, val int) { t.root = t.root.put(p, val, 0) }
func (t *kdTree) query(x1, y1, x2, y2 int) int { return t.root.query(x1, y1, x2, y2) }