2017.3.31 洞穴勘测 思考记录

        lct

            这题给我的感觉就是：指针真nm难调

                                                第一次写真蛋疼

码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define maxn 500000using namespace std; int n,m,x,y; char ss[99];struct tree{tree *ch[2],*fu;  int rev;bool isrt();void down();int getwh(){return fu->ch[0]==this?0:1;}  void set(int wh,tree *child);}lct[maxn];void tree::down(){if (rev){rev=0;swap(ch[0],ch[1]);ch[0]->rev^=1;ch[1]->rev^=1;}} void tree::set(int wh,tree *child){ch[wh]=child;if(child!=lct)child->fu=this;down();}bool tree::isrt()   //这段坑死了 {   if (fu==lct) {return 1;    }return (fu->ch[0]!=this&&fu->ch[1]!=this);}void rotate(tree *now){    tree *fu=now->fu,*ye=now->fu->fu;if (!(fu->isrt())) ye->down();//旋转的标记要提前下放     fu->down();now->down();//有序下放     int wh=now->getwh();        fu->set(wh,now->ch[wh^1]);now->set(wh^1,fu);now->fu=ye;if(!ye->isrt())ye->ch[ye->ch[0]==fu?0:1]=now; }void splay(tree *now){   if(now->isrt()) return;      for (;!(now->isrt());rotate(now))        if(!(now->fu->isrt()))now->getwh()==now->fu->getwh()?rotate(now->fu):rotate(now);}tree *access(tree *now){    tree *y=lct;  for(;now!=lct;y=now,now=now->fu)      splay(now),now->down(),now->set(1,y);//建一路路径右儿子为重     return y;}void huan(tree *a){    access(a)->rev^=1;//根打标记 splay(a);//在子splay里面氚标记，做根 }void link(tree *a,tree *b){    huan(a);//a换到根     a->fu=b;//把b放上去，相当于新建一棵splay     access(a);//a重新为根 }void cut(tree *a,tree *b){    huan(a);     //换为根     access(b);   //b放到a右边，a在第二层     splay(a);        a->ch[1]=b->fu=lct;}tree *zhao(tree *a){    access(a);    splay(a);    tree *y=a;    while(y->ch[0]!=lct)y=y->ch[0];    return y;}int main(){int i;    scanf("%d%d",&n,&m);lct->ch[0]=lct;lct->ch[1]=lct;lct->fu=lct;lct->rev=0;for (i=1;i<=n;i++){    (lct+i)->fu=(lct+i)->ch[0]=(lct+i)->ch[1]=lct;    (lct+i)->rev=0;}for (i=1;i<=m;i++){scanf("%s",ss);scanf("%d%d",&x,&y);if(ss[0]=='C')link(lct+x,lct+y);if(ss[0]=='D'){cut(lct+x,lct+y);//cout<<(lct+x)->ch[0];}if(ss[0]=='Q'){if(zhao(lct+x)==zhao(lct+y))printf("Yes\n");else printf("No\n");}}}