-
Notifications
You must be signed in to change notification settings - Fork 582
/
kd_tree.go
159 lines (140 loc) · 3.4 KB
/
kd_tree.go
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
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) }