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Black Box+POJ+treap树模板

2014-09-03 14:22 330 查看
Black Box

Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 7437Accepted: 3051
Description

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:

ADD (x): put element x into Black Box;

GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.

Let us examine a possible sequence of 11 transactions:

Example 1

N Transaction i Black Box contents after transaction Answer

(elements are arranged by non-descending)

1 ADD(3)      0 3

2 GET         1 3                                    3

3 ADD(1)      1 1, 3

4 GET         2 1, 3                                 3

5 ADD(-4)     2 -4, 1, 3

6 ADD(2)      2 -4, 1, 2, 3

7 ADD(8)      2 -4, 1, 2, 3, 8

8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8

9 GET         3 -1000, -4, 1, 2, 3, 8                1

10 GET        4 -1000, -4, 1, 2, 3, 8                2

11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8


It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.

Let us describe the sequence of transactions by two integer arrays:

1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).

2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).

The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence
we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.

Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.
Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
Sample Input
7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output
3
3
1
2

解决方案:动态的添加数据,动态求第i大的数字。直接裸treap。

code:

<pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
struct node
{
node *ch[2];
int r;///优先级
int v;///节点值
int s;
int cmp(int x)const
{
if(x==v) return -1;
else 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;
}///统计rank
};
void rotate(node* & rt,int d)
{
node *k=rt->ch[d^1];
rt->ch[d^1]=k->ch[d];
k->ch[d]=rt;
rt->maintain();
k->maintain();///必须先更新rt再更新k
rt=k;
}///left rotate and right rotate

void insert(node * & rt,int x)
{
if(rt==NULL)
{
rt=new node();
rt->ch[0]=rt->ch[1]=NULL;
rt->v=x;
rt->s=1;
rt->r=rand();///建立新的节点,插入数字
}
else
{
int d=(x<=rt->v?0:1);
insert(rt->ch[d],x);///选择合适的位置插入
if((rt->ch[d]->r)>rt->r)
rotate(rt,d^1);///左孩子优先级大右旋,右孩纸优先级大左旋
}
rt->maintain();///儿子节点发生改变,其父节点也要更新
}
int kth(node *rt,int k)
{
if(rt==NULL||k<=0||k>rt->s) return 0;
int s=(rt->ch[0]==NULL?0:rt->ch[0]->s);
if(s+1==k) return rt->v;
else if(s>=k) return kth(rt->ch[0],k);
else return kth(rt->ch[1],k-s-1);

}///查询第k大的数字
int v[30004];
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k))
{
for(int i=1; i<=n; i++)
{
scanf("%d",&v[i]);
}
node *root=NULL;
int now=1;
for(int i=1; i<=k; i++)
{
int x;
scanf("%d",&x);
for(; now<=x; now++)
insert(root,v[now]);
printf("%d\n",kth(root,i));
}

}
return 0;
}


[/code]


                                            
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