BZOJ 2733 [HNOI2012] 永无乡 Treap

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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 100005using 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[N];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[N];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;}

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