poj-1442 Black Box(Treap)

题目链接：

Black Box

题意：

给一个序列,m个询问,每个询问是求前x个数中的第i小是多少;

思路：

Treap的入门题目;Treap能实现STL的set实现不了的功能,如名次树(rank tree)rank tree 支持两种新操作,一个是找出第k小元素,一个是找出x的的rank;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('\n');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9+10;
const int N=1e6+10;
const int maxn=1e3+20;
const double eps=1e-12;

int a
;
struct Treap
{
Treap *ch[2];
int siz,key,fix;
Treap(int key)
{
siz=1;
fix=rand();
this->key=key;
ch[0]=ch[1]=NULL;
}
int cmp(int x) const
{
if(x==key)return -1;
return x<key? 0:1;
}
void maintain()
{
siz=1;
if(ch[0]!=NULL)siz+=ch[0]->siz;
if(ch[1]!=NULL)siz+=ch[1]->siz;
}
};

void Rotate(Treap* &o,int d)
{
Treap* k=o->ch[d^1];
o->ch[d^1]=k->ch[d];
k->ch[d]=o;
o->maintain();
k->maintain();
o=k;
}

void insert(Treap* &o,int x)
{
if(o==NULL)o=new Treap(x);
else
{
int d=(x<o->key? 0:1);
insert(o->ch[d],x);
if(o->ch[d]->fix > o->fix)Rotate(o,d^1);
}
o->maintain();
}
int  getans(Treap* &o,int x)
{
//cout<<o->key<<"*****\n";
if(o==NULL||x<=0||x> o->siz)return 0;
int s=(o->ch[0] == NULL? 0: o->ch[0]->siz);
if(x==s+1)return o->key;
else if(x<=s)return getans(o->ch[0],x);
else return getans(o->ch[1],x-s-1);
}
void removetree(Treap* &o)
{
if(o->ch[0]!=NULL)removetree(o->ch[0]);
if(o->ch[1]!=NULL)removetree(o->ch[1]);
delete o;
o=NULL;
}

int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
For(i,1,n)read(a[i]);
int pre=1,x;
Treap* root=NULL;
For(i,1,m)
{
read(x);
for(int j=pre;j<=x;j++)insert(root,a[j]);
printf("%d\n",getans(root,i));
pre=x+1;
}
removetree(root);
}
return 0;
}