A+B问题简单的题解
2023-07-28 17:09:11
发布于：河北
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终于可以写一份A+B这么难的题的题解了。
咦？竟然没有人写LCT的题解？
Link-Cut Tree 会很伤心的！
ORZ为了不让LCT伤心于是我来一份LCT的A+B题解吧！
送上代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstring>
using namespace std;
struct node {
int data,rev,sum;
node *son[2],*pre;
bool judge();
bool isroot();
void pushdown();
void update();
void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x){
node *now=lct+ ++top;
now->data=x;
now->pre=now->son[1]=now->son[0]=lct;
now->sum=0;
now->rev=0;
return now;
}
bool node::judge(){return pre->son[1]==this;}
bool node::isroot(){
if(pre==lct)return true;
return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown(){
if(this==lct||!rev)return;
swap(son[0],son[1]);
son[0]->rev^=1;
son[1]->rev^=1;
rev=0;
}
void node::update(){sum=son[1]->sum+son[0]->sum+data;}
void node::setson(node *child,int lr){
this->pushdown();
child->pre=this;
son[lr]=child;
this->update();
}
void rotate(node *now){
node *father=now->pre,*grandfa=father->pre;
if(!father->isroot()) grandfa->pushdown();
father->pushdown();now->pushdown();
int lr=now->judge();
father->setson(now->son[lr^1],lr);
if(father->isroot()) now->pre=grandfa;
else grandfa->setson(now,father->judge());
now->setson(father,lr^1);
father->update();now->update();
if(grandfa!=lct) grandfa->update();
}
void splay(node *now){
if(now->isroot())return;
for(;!now->isroot();rotate(now))
if(!now->pre->isroot())
now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now){
node *last=lct;
for(;now!=lct;last=now,now=now->pre){
splay(now);
now->setson(last,1);
}
return last;
}
void changeroot(node *now){
access(now)->rev^=1;
splay(now);
}
void connect(node *x,node *y){
changeroot(x);
x->pre=y;
access(x);
}
void cut(node *x,node *y){
changeroot(x);
access(y);
splay(x);
x->pushdown();
x->son[1]=y->pre=lct;
x->update();
}
int query(node *x,node *y){
changeroot(x);
node *now=access(y);
return now->sum;
}
int main(){
scanf("%d%d",&a,&b);
node *A=getnew(a);
node *B=getnew(b);
//连边 Link
connect(A,B);
//断边 Cut
cut(A,B);
//再连边orz Link again
connect(A,B);
printf("%d\n",query(A,B));
return 0;
}
同志们，我们就从这一题开始，向着大犇的路进发！
任何一个伟大的思想，都有一个微不足道的开始！
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