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Black Box(POJ 1442·TREAP实现)

2024-07-15 07:49:57   222人阅读

传送门：http://poj.org/problem?id=1442

 

Black Box
Time Limit: 1000MS Memory Limit: 10000K
   

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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

Source

Northeastern Europe 1996
 
 
第二棵treap。。总体感觉treap非常好写，我学treap是为了防止有的题卡splay。。感觉应该差不多了QAQ...
这题没什么做法。。
Codes:
 1 #include<set> 2 #include<ctime> 3 #include<queue> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std;10 const int N = 100100;11 #define L(i) (T[i].s[0])12 #define R(i) (T[i].s[1])13 #define For(i,n) for(int i=1;i<=n;i++)14 #define Rep(i,l,r) for(int i=l;i<=r;i++)15 16 struct treap{17     int size,s[2],v,pri;18     void Sets(int x,int y){19         size = 1;v = x;pri = y;20     }21 }T[N];22 23 int n,m,A[N],size,Lim,now,level = 0;24 int tot,root;25 int read(){26     char ch = getchar(); int num = 0 , q = 1;27     while(ch>‘9‘||ch<‘0‘){28         if(ch==‘-‘) q = -1;29         ch = getchar();30     }31     while(ch>=‘0‘&&ch<=‘9‘){32         num = num * 10 + ch - ‘0‘;33         ch = getchar();34     }35     return num * q;36 }37 38 void Update(int i){39     T[i].size = T[L(i)].size + T[R(i)].size + 1;40 }41 42 void Rot(int &y,int f){43     int x = T[y].s[!f];44     T[y].s[!f] = T[x].s[f];45     T[x].s[f]  = y;46     Update(y);Update(x);47     y = x;48 }49 50 void Insert(int &i,int val){51     if(!i){52         T[i=++tot].Sets(val,rand());53         return;54     }55     int f = T[i].v > val;56     Insert(T[i].s[!f],val);57     if(T[T[i].s[!f]].pri > T[i].pri) Rot(i,f);58     else                             Update(i);59 }60 61 int Rank(int i,int kth){62     if(T[L(i)].size + 1 == kth) return i;63     else if(T[L(i)].size >=kth) return Rank(L(i),kth);64     else return Rank(R(i),kth - T[L(i)].size - 1);65 }66 67 int main(){68     srand(time(NULL));69     n = read(); m = read();70     For(i,n) A[i] = read();71     For(i,m) {72         Lim = read();73         Rep(i,now+1,Lim) Insert(root,A[i]); now = Lim;74         level++;printf("%d\n",T[Rank(root,level)].v);75     }76     return 0;77 }

 


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