poj 1442 名次树

这回要求的是第k小的元素，

参考了ljl大神的模板，orz

 

//insert 插入
//remove 删除
//_find 查找
//kth 返回root为根的树中第k小的元素
//treap插入、删除、查询时间复杂度均为O(logn)
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cstdio>
using namespace std;
int a[300010],n,m;
struct Node
    {
        Node *ch[2];
        int r,v,s;
        Node(int v):v(v)
            {s=1;ch[0]=ch[1]=NULL;r=rand();}
        int cmp(int x) const{
            if(x==v)return(-1);
            return (x<v)?(0):(1);
        }
        void maintain(){
            s=1;
            if(ch[0]!=NULL)s+=ch[0]->s;
            if(ch[1]!=NULL)s+=ch[1]->s;
        }
    };
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;
    }
void insert(Node* &o,int x)
    {
        if(o==NULL) o=new Node(x);
        else {
            int d=(x<o->v)?0:1;
            insert(o->ch[d],x);
            if((o->r)<(o->ch[d]->r))
                rotate(o,d^1);
        }
        o->maintain();
    }
void remove(Node* &o,int x)
    {
        int d=o->cmp(x);
        if(d==-1){
            Node* u=o;
            if(o->ch[0]==NULL){o=o->ch[1];delete u;}else
            if(o->ch[1]==NULL){o=o->ch[0];delete u;}else {
                int c=(o->ch[0])>(o->ch[1])?(1):(0);
                rotate(o,c);
                remove(o->ch[c],x);
            }
        }
        else remove(o->ch[d],x);
        if(o!=NULL) o->maintain();
    }
int find(Node* &o,int x)
    {
        while(o!=NULL){
            int d=o->cmp(x);
            if(d==-1)
                return 1;
            else
                o=o->ch[d];
        }
        return 0;
    }
int kth(Node* o,int k)
    {
        while(o!=NULL){
            int lchsize=(o->ch[0]!=NULL)?
                        (o->ch[0]->s):0;
            if(lchsize+1==k)
                return o->v;
            else {
                int d=(lchsize<k);
                o=o->ch[d];
                k-=d*(lchsize+1);
            }
        }
        return 0;
    }

int main()
{
    Node* root=NULL;
    int u,ans,p=0;
    srand(time(0));

    cin>>m>>n;
    for(int i=0;i<m;i++)
        cin>>a[i];
    for(int i=1;i<=n;i++)
    {
        cin>>u;
        while(p<u)
            insert(root,a[p++]);
        ans=kth(root,i);
        cout<<ans<<endl;
    }
    return 0;
}