LCT模板(指针版)
本来是想做THUWC2017的泰勒展开xLCT题的……
然后觉得数组写很麻烦……
然后就决定挑战指针版……
然后写得全是BUG……
与BUG鏖战三千年后,有了这个指针版LCT板子!
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
#define enter putchar('\n')
using namespace std;
typedef long long ll;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
typedef long double ldb;
const int N = 300005;
int n, m;
struct node {
node *fa, *ch[2];
int val, sum;
bool rev;
node(){
fa = ch[0] = ch[1] = NULL;
val = sum = 0;
rev = 0;
}
bool which(){
return fa->ch[1] == this;
}
bool isroot(){
return fa == NULL || (fa->ch[0] != this && fa->ch[1] != this);
}
void upt(){
sum = val;
if(ch[0] != NULL) sum ^= ch[0]->sum;
if(ch[1] != NULL) sum ^= ch[1]->sum;
}
void pushdown(){
if(!rev) return;
swap(ch[0], ch[1]);
if(ch[0] != NULL) ch[0]->rev ^= 1;
if(ch[1] != NULL) ch[1]->rev ^= 1;
rev = 0;
}
} s[N];
void rotate(node *u){
node *v = u->fa, *w = v->fa, *b = u->ch[!u->which()];
if(!v->isroot()) w->ch[v->which()] = u;
u->which() ? (u->ch[0] = v, v->ch[1] = b) : (u->ch[1] = v, v->ch[0] = b);
u->fa = w, v->fa = u;
if(b != NULL) b->fa = v;
v->upt();
}
void splay(node *u){
static node *stk[N];
int top;
stk[top = 1] = u;
while(!stk[top]->isroot()) stk[top + 1] = stk[top]->fa, top++;
while(top) stk[top--]->pushdown();
while(!u->isroot()){
if(!u->fa->isroot()){
if(u->which() == u->fa->which()) rotate(u->fa);
else rotate(u);
}
rotate(u);
}
u->upt();
}
void access(node *u){
node *v = NULL;
while(u != NULL){
splay(u);
u->ch[1] = v;
u->upt();
v = u;
u = u->fa;
}
}
void makeroot(node *u){
access(u);
splay(u);
u->rev ^= 1;
}
node *findroot(node *u){
access(u);
splay(u);
while(u->pushdown(), u->ch[0] != NULL)
u = u->ch[0];
splay(u);
return u;
}
void link(node *u, node *v){
if(findroot(u) == findroot(v)) return;
makeroot(v);
v->fa = u;
}
void cut(node *u, node *v){
makeroot(u);
access(v);
splay(v);
if(v->ch[0] == u)
v->ch[0] = u->fa = NULL;
}
int query(node *u, node *v){
makeroot(u);
access(v);
splay(v);
return v->sum;
}
void change(node *u, int x){
splay(u);
u->val = x;
u->upt();
}
int main(){
read(n), read(m);
for(int i = 1; i <= n; i++)
read(s[i].val), s[i].upt();
int op, x, y;
while(m--){
read(op), read(x), read(y);
if(op == 0) write(query(s + x, s + y)), enter;
else if(op == 1) link(s + x, s + y);
else if(op == 2) cut(s + x, s + y);
else change(s + x, y);
}
return 0;
}
本文作者:胡小兔
博客地址:http://rabbithu.cnblogs.com
博客地址:http://rabbithu.cnblogs.com