「BZOJ 5020」在美妙的数学王国中畅游-Link-Cut-Tree

维护一个森林,每个点有一个函数 $f$,要求支持连接两个点,断开两个点,修改某个点的函数,询问路径上函数 $f(x)$ 的和。

链接

BZOJ 5020

题解

考虑麦克劳林展开
$$f(x)=\sum_{n = 0} ^ {\infty}f ^ {(n)} (0) \frac{x ^ n}{n!}$$

于是我们只需要对每个点维护前 $k$ 项的系数就可以了,$k$ 并不会很大,大概 $11$ 就可以保证精度。

对于 $f(x) = \sin(ax + b)$ 时
$$\begin{aligned}f'(0) &= a \cdot \cos(b) \\ f ^ {(2)}(0) &= - a ^ 2 \cdot \sin(b) \\ f ^ {(3)}(0) &= - a ^ 3 \cdot \cos(b) \\ f ^ {(4)}(0) &= a ^ 4 \cdot \sin(b)\end{aligned}$$
至此已经出现循环,后面的直接递推就好了。

对于 $f(x) = e ^ {ax + b}$ 时
$$f ^ {(i)}(0) = a ^ i \cdot e ^ b$$

对于 $f(x) = ax + b$ 时
$$\begin{aligned}f'(0) &= b \\ f''(0) &= a\end{aligned}$$
其他均为 $0$

然后用 LCT 维护每个系数的和即可。

时间复杂度 $O(kn \log n)$

代码

小常数 LCT + 循环展开效果很好啊,BZOJ rk1,比 rk2 快了 6s,23333….

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#include <bits/stdc++.h>
namespace {
inline char read() {
static const int IN_LEN = 1 << 18 | 1;
static char buf[IN_LEN], *s, *t;
return (s == t) && (t = (s = buf) + fread(buf, 1, IN_LEN, stdin)),
s == t ? -1 : *s++;
}
template <typename T>
inline void read(T &x) {
static char c;
static bool iosig;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
iosig |= c == '-';
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
iosig && (x = -x);
}
inline int read(char *buf) {
register int s = 0;
register char c;
while (c = read(), isspace(c) && c != -1)
;
if (c == -1) {
*buf = 0;
return -1;
}
do
buf[s++] = c;
while (c = read(), !isspace(c) && c != -1);
buf[s] = 0;
return s;
}
inline void read(char &c) {
while (c = read(), isspace(c) && c != -1)
;
}
inline void read(double &t) {
static char c;
static bool iosig;
register int x = 0;
for (c = read(), iosig = false; !isdigit(c); c = read()) {
if (c == -1) return;
c == '-' ? iosig = true : 0;
}
for (x = 0; isdigit(c); c = read()) x = x * 10 + (c ^ '0');
if (c == '.') {
register long long y = 0, cnt = 1;
for (c = read(); isdigit(c); c = read())
y = y * 10 + (c ^ '0'), cnt *= 10;
t = x + (double)y / cnt;
} else {
t = x;
}
iosig ? t = -t : 0;
}
const int OUT_LEN = 1 << 18 | 1;
char obuf[OUT_LEN], *oh = obuf;
inline void print(char c) {
(oh == obuf + OUT_LEN) && (fwrite(obuf, 1, OUT_LEN, 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');
}
}
inline void print(const char *s) {
for (; *s; s++) print(*s);
}
const double EPS = 1e-9;
inline int sign(const double x) { return (x > EPS) - (x < -EPS); }
inline void print(double x) {
register int sig = sign(x);
if (sig == 0) {
print('0');
return;
}
(sig == -1) && (print('-'), x = -x);
print((int)x);
x = x - (int)x;
print('.');
static char buf[21], cnt;
x *= 1000000000;
register int t = round(x);
for (cnt = 0; t; t /= 10) buf[++cnt] = t % 10 | 48;
for (register int i = 9; i > cnt; i--) print('0');
while (cnt) print((char)buf[cnt--]);
}
struct InputOutputStream {
~InputOutputStream() { fwrite(obuf, 1, oh - obuf, stdout); }
template <typename T>
inline InputOutputStream &operator>>(T &x) {
read(x);
return *this;
}
template <typename T>
inline InputOutputStream &operator<<(const T &x) {
print(x);
return *this;
}
} io;
const int MAXN = 100000 + 9;
const int MAX_EXP = 11;
struct Node *null;
struct Node {
double f[MAX_EXP], sum[MAX_EXP];
Node *c[2], *fa, *top;
bool rev;
inline bool relation() { return this == fa->c[1]; }
inline void reverse() { rev ^= 1, std::swap(c[0], c[1]); }
inline void pushDown() {
rev && (c[0]->reverse(), c[1]->reverse(), rev = false);
}
inline void maintain() {
#define opt(i) sum[i] = f[i] + c[0]->sum[i] + c[1]->sum[i]
opt(0);
opt(1), opt(2), opt(3), opt(4), opt(5);
opt(6), opt(7), opt(8), opt(9), opt(10);
#undef opt
}
inline void rotate(register bool f) {
register Node *o = fa;
top = o->top;
o->pushDown();
pushDown();
(fa = o->fa)->c[o->relation()] = this;
(o->c[f] = c[!f])->fa = o;
(c[!f] = o)->fa = this;
o->maintain();
}
inline void splay() {
register bool f;
for (pushDown(); fa != null;) {
(f = relation(), fa->fa == null)
? rotate(f)
: (f == fa->relation() ? (fa->rotate(f), rotate(f))
: (rotate(f), rotate(!f)));
}
maintain();
}
inline void expose() {
splay();
if (c[1] != null) {
c[1]->top = this;
c[1]->fa = null;
c[1] = null;
maintain();
}
}
inline bool splice() {
splay();
if (top == null) return false;
top->expose();
top->c[1] = this;
fa = top;
top = null;
fa->maintain();
return true;
}
inline void access() {
for (expose(); splice();)
;
}
inline void evert() { access(), splay(), reverse(); }
} pool[MAXN];
inline void link(register Node *u, register Node *v) {
u->evert();
u->top = v;
}
inline void cut(register Node *u, register Node *v) {
u->expose();
v->expose();
if (u->top == v) u->top = null;
if (v->top == u) v->top = null;
}
inline void modify(register Node *p, const int cmd, const double a,
const double b) {
memset(p->f, 0, sizeof(p->f));
p->splay();
switch (cmd) {
case 1: {
// sin(b), cos(b) * a, -sin(b) * a ^ 2
// -cos(b) * a ^ 3, sin(b) * a ^ 4
register double expN = a;
const double SIN_B = sin(b), COS_B = cos(b);
p->f[0] = SIN_B;
p->f[1] = COS_B * expN;
expN *= a;
p->f[2] = -(SIN_B * expN) / 2;
expN *= a;
p->f[3] = -(COS_B * expN) / 6;
expN *= a;
p->f[4] = (SIN_B * expN) / 24;
#define opt(i) p->f[i] = p->f[i - 4] * expN / (i - 3) / (i - 2) / (i - 1) / i
opt(5), opt(6), opt(7), opt(8), opt(9), opt(10);
#undef opt
break;
}
case 2: {
// a ^ i \cdot e ^ b
p->f[0] = exp(b);
#define opt(i) p->f[i] = p->f[i - 1] * a / i
opt(1), opt(2), opt(3), opt(4), opt(5);
opt(6), opt(7), opt(8), opt(9), opt(10);
#undef opt
break;
}
case 3: {
p->f[0] = b, p->f[1] = a;
break;
}
}
p->maintain();
}
inline void query(register Node *u, register Node *v, register double x) {
register double expN = 1, ans = 0;
u->evert(), v->access(), v->splay();
register Node *p = v;
for (; p->c[0] != null; p = p->c[0]) p->pushDown();
if (p != u) {
io << "unreachable\n";
return;
}
for (register int i = 0; i < MAX_EXP; i++) {
ans += expN * v->sum[i];
expN *= x;
}
io << ans << '\n';
}
inline void solve() {
null = pool;
null->c[0] = null->c[1] = null;
null->fa = null;
null->top = null;
register int n, m;
static char cmd, buf[21];
io >> n >> m >> buf;
for (register int i = 1, type; i <= n; i++) {
pool[i].c[0] = pool[i].c[1] = null;
pool[i].fa = null;
pool[i].top = null;
register double a, b;
io >> type >> a >> b;
modify(pool + i, type, a, b);
}
while (m--) {
io >> cmd;
switch (cmd) {
case 'a': {
register int u, v;
io >> u >> v;
link(pool + u + 1, pool + v + 1);
break;
}
case 'd': {
register int u, v;
io >> u >> v;
cut(pool + u + 1, pool + v + 1);
break;
}
case 'm': {
register int c, f;
register double a, b;
io >> c >> f >> a >> b;
modify(pool + c + 1, f, a, b);
break;
}
case 't': {
register int u, v;
register double x;
io >> u >> v >> x;
query(pool + u + 1, pool + v + 1, x);
break;
}
}
}
}
} // namespace
int main() {
// freopen("math.in", "r", stdin);
// freopen("math.out", "w", stdout);
solve();
return 0;
}

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