「BZOJ-2631」tree-Link-Cut Tree

一棵n个点的树,每个点的初始权值为1。对于这棵树有q个操作,每个操作为以下四种操作之一:

+ $u$ $v$ $c$:将u到v的路径上的点的权值都加上自然数c;

- $u_1$ $v_1$ $u_2$ $v_2$:将树中原有的边 $(u_1,v_1)$ 删除,加入一条新边 $(u_2,v_2)$,保证操作完之后仍然是一棵树;

* $u$ $v$ $c$:将 $u$ 到 $v$ 的路径上的点的权值都乘上自然数 $c$;

/ $u$ $v$:询问 $u$ 到 $v$ 的路径上的点的权值和,求出答案对于 $51061$ 的余数。

链接

BZOJ-2631

题解

模板题,还需要题解?
注意: 此题只需要 unsigned int 不需要 long long

论三目运算符与 ifelse 的常数差距……直接拿下了BZOJ rank1

经过三道题,看来我的 LCT 常数确实较小。

代码

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/*
* created by xehoth on 21-03-2017
*/
#include <bits/stdc++.h>
typedef unsigned int uint;
const int MAXN = 100010;
const int MOD = 51061;
struct Node *null;
struct Node {
Node *c[2], *fa;
bool rev;
Node *top;
uint sum, add, mul, size, val;
Node() : mul(1), sum(1), val(1), size(1), fa(null) {
c[0] = c[1] = null;
}
inline void cover(int m, int a) {
val = (val * m + a) % MOD;
sum = (sum * m + a * size) % MOD;
add = (add * m + a) % MOD;
mul = (mul * m) % MOD;
}
inline void maintain() {
sum = (c[0]->sum + c[1]->sum + val) % MOD;
size = (c[0]->size + c[1]->size + 1) % MOD;
}
inline void reverse() {
rev ^= 1, std::swap(c[0], c[1]);
}
inline void pushDown() {
rev ? c[0]->reverse(), c[1]->reverse(), rev = false : 0;
mul != 1 || add != 0 ? c[0]->cover(mul, add),
c[1]->cover(mul, add), mul = 1, add = 0 : 0;
}
inline bool relation() {
return this == fa->c[1];
}
inline void rotate(bool f) {
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();
}
/*
我也不想这么写,
我也很绝望啊,
然而三目运算符常数小啊......
*/
inline void splay() {
Node *o = fa;
bool f;
for (pushDown(); o != null; o = fa) {
o->fa == null ? rotate(o->c[1] == this) :
((f = o->c[1] == this) == (o->fa->c[1] == o)
? (o->rotate(f), rotate(f)) : (rotate(f), rotate(!f)));
}
maintain();
}
inline void expose(Node *p = null) {
splay();
if (c[1] != null)
c[1]->top = this, c[1]->fa = null;
(c[1] = p)->fa = this;
maintain();
}
inline Node *access() {
Node *x = this;
for (x->expose(); x->top; x = x->top)
x->top->expose(x);
return x;
}
inline void evert() {
access(), splay(), reverse();
}
inline void link(Node *f) {
Node *x = access();
x->reverse(), x->top = f;
}
inline void cut(Node *y) {
Node *x = this;
x->expose(), y->expose();
if (x->top == y) x->top = NULL;
if (y->top == x) y->top = NULL;
}
inline Node *findRoot() {
Node *f = this;
f->access(), f->splay();
while (f->pushDown(), f->c[0] != null) f = f->c[0];
f->splay();
return f;
}
inline void split(Node *v) {
v->evert(), access(), splay();
}
} pool[MAXN];
inline void init() {
null = pool, null->fa = null;
null->sum = null->val = null->size = null->add = 0;
null->mul = 1;
}
inline char nextChar() {
static const int IN_LEN = 1000000;
static char buf[IN_LEN], *s, *t;
if (s == t) {
t = (s = buf) + fread(buf, 1, IN_LEN, stdin);
if (s == t) return -1;
}
return *s++;
}
inline int read() {
static int x = 0;
static char c;
for (x = 0, c = nextChar(); !isdigit(c); c = nextChar());
for (; isdigit(c); c = nextChar())
x = (x + (4 * x) << 1) + (c ^ '0');
return x;
}
const int OUT_LEN = 10000000;
char obuf[OUT_LEN], *oh = obuf;
template<class T>
inline void print(T x) {
static int buf[30], cnt;
if (x == 0) {
*oh++ = '0';
} else {
if (x < 0) *oh++ = '-', x = -x;
register int cnt = 0;
for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48;
while (cnt) *oh++ = buf[cnt--];
}
}
template<class T>
inline void println(T x) {
print(x), *oh++ = '\n';
}
inline void flush() {
fwrite(obuf, 1, oh - obuf, stdout);
}
int n, q;
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in", "r", stdin);
#endif
init();
n = read(), q = read();
for (register int i = 1; i <= n; i++) pool[i] = Node();
char s;
register int x, y, z;
for (register int i = 1; i < n; i++) (pool + read())->link(pool + read());
for (register int i = 1; i <= q; i++) {
s = nextChar(), x = read(), y = read();
switch (s) {
case '+':
z = read(), (pool + x)->split(pool + y);
(pool + x)->cover(1, z);
break;
case '-':
(pool + x)->cut(pool + y),
x = read(), y = read(),
(pool + x)->link(pool + y);
break;
case '*':
z = read(), (pool + x)->split(pool + y);
(pool + x)->cover(z, 0);
break;
case '/':
(pool + x)->split(pool + y), println((pool + x)->sum);
break;
}
}
flush();
return 0;
}
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