「BZOJ 4311」向量-线段树分治 + 凸包

维护一个向量集合,支持:

  1. 插入一个向量 $(x, y)$;
  2. 删除插入的第 $i$ 个向量;
  3. 查询当前集合与 $(x, y)$ 点积的最大值是多少,如果当前是空集输出 $0$。

链接

BZOJ 4311

题解

$(a, b) \cdot (x, y) = ax + by$,令 $c = ax + by$,则 $y = - \frac {a} {b} x + \frac {c} {b}$,由于 $(a, b)$ 是询问给出的定值,要使 $c$ 最大,即截距最大,那么答案一定在上凸壳取到。

由于有删除操作,考虑用线段树分治来维护,以时间建立线段树,线段树上每个节点维护一个上凸壳,询问时显然我们可以直接在每一段凸壳上三分,但还有一种做法,我们先把询问按 $- \frac {a} {b}$ 从大到小排序,那么决策就是单调的了,直接在凸壳上维护一个指针扫过去就可以了。

时间复杂度 $O(n \log n)$。

代码

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
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
/**
* Copyright (c) 2016-2018, xehoth
* All rights reserved.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
* http://www.apache.org/licenses/LICENSE-2.0
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
* 「BZOJ 4311」向量 26-02-2018
* 线段树分治 + 凸包
* @author xehoth
*/
#include <algorithm>
#include <cctype>
#include <cstdio>
#include <iostream>
#include <vector>
struct InputOutputStream {
enum { SIZE = 1 << 18 | 1 };
char ibuf[SIZE], *s, *t, obuf[SIZE], *oh;
InputOutputStream() : s(), t(), oh(obuf) {}
~InputOutputStream() { fwrite(obuf, 1, oh - obuf, stdout); }
inline char read() {
return (s == t) && (t = (s = ibuf) + fread(ibuf, 1, SIZE, stdin)),
s == t ? -1 : *s++;
}
template <typename T>
inline InputOutputStream &operator>>(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return *this;
iosig |= c == '-';
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig && (x = -x);
return *this;
}
inline void print(char c) {
(oh == obuf + SIZE) && (fwrite(obuf, 1, SIZE, stdout), oh = obuf);
*oh++ = c;
}
template <typename T>
inline void print(T x) {
static int buf[21], cnt;
if (x != 0) {
(x < 0) && (print('-'), x = -x);
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 | 48;
while (cnt) print((char)buf[cnt--]);
} else {
print('0');
}
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
} io;
const int MAXN = 200000 + 9;
struct Point {
int x, y;
Point() {}
Point(int x, int y) : x(x), y(y) {}
inline Point operator-(const Point &p) const {
return Point(x - p.x, y - p.y);
}
inline long long operator*(const Point &p) const {
return (long long)x * p.y - (long long)y * p.x;
}
inline bool operator<(const Point &p) const {
return x < p.x || (x == p.x && y < p.y);
}
};
inline long long dot(const Point &a, const Point &b) {
return (long long)a.x * b.x + (long long)a.y * b.y;
}
struct Node *null;
char *cur;
struct Node {
static const int NODE_SIZE;
Node *lc, *rc;
std::vector<const Point *> c;
int pos;
inline void insert(const Point *p) {
while (c.size() > 1 &&
(*p - *c[c.size() - 1]) * (*p - *c[c.size() - 2]) <= 0)
c.pop_back();
c.push_back(p);
}
inline long long query(const Point *p) {
if (c.empty()) return 0;
while (pos < (int)c.size() - 1 &&
dot(*p, *c[pos + 1]) >= dot(*p, *c[pos]))
pos++;
return dot(*p, *c[pos]);
}
Node() : lc(null), rc(null), c(), pos() {}
inline void *operator new(size_t) { return cur += NODE_SIZE; }
};
const int Node::NODE_SIZE = sizeof(Node);
char pool[MAXN * Node::NODE_SIZE * 4];
struct Data {
Point p;
int l, r;
} d[MAXN], *que[MAXN], *id[MAXN];
int iCnt, eCnt, qCnt;
inline bool cmpX(const Data *a, const Data *b) { return a->p < b->p; }
inline bool cmpQ(const Data *a, const Data *b) { return a->p * b->p < 0; }
class SegmentTree {
public:
SegmentTree(const int n) : root(null), n(n) { build(root, 1, n); }
inline void insert(int l, int r, const Point *p) {
insert(root, 1, n, l, r, p);
}
inline long long query(int k, const Point *p) {
return query(root, 1, n, k, p);
}
private:
Node *root;
int n;
void build(Node *&p, int l, int r) {
p = new Node;
if (l == r) return;
int mid = (l + r) >> 1;
build(p->lc, l, mid);
build(p->rc, mid + 1, r);
}
void insert(Node *p, int l, int r, int s, int t, const Point *pts) {
if (s <= l && t >= r) {
p->insert(pts);
return;
}
int mid = (l + r) >> 1;
if (s <= mid) insert(p->lc, l, mid, s, t, pts);
if (t > mid) insert(p->rc, mid + 1, r, s, t, pts);
}
long long query(Node *p, int l, int r, int k, const Point *pts) {
long long ans = p->query(pts);
if (l == r) return ans;
int mid = (l + r) >> 1;
return k <= mid ? std::max(ans, query(p->lc, l, mid, k, pts))
: std::max(ans, query(p->rc, mid + 1, r, k, pts));
}
};
long long ans[MAXN];
int main() {
// freopen("sample/1.in", "r", stdin);
cur = pool;
null = (Node *)pool;
null->lc = null;
null->rc = null;
null->pos = 0;
null->c = std::vector<const Point *>();
int n;
io >> n;
for (int i = 1, cmd, t; i <= n; i++) {
io >> cmd;
switch (cmd) {
case 1: {
io >> d[i].p.x >> d[i].p.y;
d[i].l = i;
d[i].r = n;
id[++iCnt] = d + i;
break;
}
case 2: {
io >> t;
id[t]->r = i - 1;
d[i].r = -1;
break;
}
case 3: {
io >> d[i].p.x >> d[i].p.y;
d[i].l = i;
que[++qCnt] = d + i;
break;
}
}
}
SegmentTree seg(n);
iCnt = 0;
for (int i = 1; i <= n; i++)
if (d[i].r > 0) id[++iCnt] = d + i;
std::sort(id + 1, id + iCnt + 1, cmpX);
for (int i = 1; i <= iCnt; i++) seg.insert(id[i]->l, id[i]->r, &id[i]->p);
std::sort(que + 1, que + qCnt + 1, cmpQ);
for (int i = 1; i <= qCnt; i++)
ans[que[i]->l] = seg.query(que[i]->l, &que[i]->p);
for (int i = 1; i <= n; i++)
if (d[i].r == 0) io << ans[i] << '\n';
return 0;
}
分享到