BZOJ 2733 [HNOI2012] 永无乡 Treap

题目大意：给出n个只有一个元素的集合，有以下操作：集合合并，查集合第k小。

查询集合内第k小，平衡树可以搞。

合并怎么办？

启发式合并。暴力拆解一个size比较小的，保证每个点至多被拆log(n)次。

如果俩点在一个集合里就不能再合并了否则会RE

#include <cstdio>
#include <cstdlib>
#define N 100005
using namespace std;
struct Point {
int ord,val;
Point(int x=0,int y=0):ord(x),val(y){}
bool operator < (const Point &rhs) const {return val<rhs.val;}
bool operator == (const Point &rhs) const {return val==rhs.val;}
};
struct Node {
Node* ch[2];
int s,r;
Point v;
int cmp(Point x) {
if(x==v) return -1;
return x < v ? 0 : 1;
}
void maintain() {
s=ch[0]->s+ch[1]->s+1;
return ;
}
Node(Point);
}*null=new Node(Point(0,0)),*treap
;
Node :: Node(Point x):v(x){ r=rand(); s=null ? 1 : 0; ch[0]=ch[1]=null;}
void Rotate(Node*& o,int d) {
Node* k=o->ch[d^1];
o->ch[d^1]=k->ch[d];
k->ch[d]=o;
o->maintain(); k->maintain();
o=k;
return ;
}
void Insert(Node*& o,Point x) {
if(o==null) {o=new Node(x); return ;}
int d=o->cmp(x);
Insert(o->ch[d],x);
if(o->ch[d]->r > o->r) Rotate(o,d^1);
o->maintain();
return ;
}
void Merge(Node*& o,Node*& x) {
if(o==null) return ;
Merge(o->ch[0],x); Merge(o->ch[1],x);
Insert(x,o->v);
delete o;
o=null;
return ;
}
int Kth(Node* o,int x) {
if(x > o->s || x<1 || o==null) return -1;
if(o->ch[0]->s+1 == x) return o->v.ord;
if(o->ch[0]->s+1 > x) return Kth(o->ch[0],x);
return Kth(o->ch[1],x-o->ch[0]->s-1);
}
int pa
;
int root(int x) {return pa[x]==x ? pa[x] : pa[x]=root(pa[x]);}
int main() {
null->ch[0]=null->ch[1]=null;
int n,m,x,y;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%d",&x) , treap[i]=new Node(Point(i,x)) , pa[i]=i;
while(m--) {
scanf("%d%d",&x,&y);
int pa_x=root(x),pa_y=root(y);
if(pa_x==pa_y) continue;
if(treap[pa_x]->s > treap[pa_y]->s) Merge(treap[pa_y],treap[pa_x]) , pa[pa_y]=pa_x;
else Merge(treap[pa_x],treap[pa_y]) , pa[pa_x]=pa_y;
}
scanf("%d",&m);
while(m--) {
char mode[2];
scanf("%s%d%d",mode,&x,&y);
int pa_x=root(x);
if(mode[0]=='Q') printf("%d\n",Kth(treap[pa_x],y));
if(mode[0]=='B') {
int pa_y=root(y);
if(pa_x==pa_y) continue;
if(treap[pa_x]->s > treap[pa_y]->s) Merge(treap[pa_y],treap[pa_x]) , pa[pa_y]=pa_x;
else Merge(treap[pa_x],treap[pa_y]) , pa[pa_x]=pa_y;
}
}
return 0;
}