Black Box(POJ 1442·TREAP实现)

传送门：http://poj.org/problem?id=1442

Black Box

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

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 }

Black Box(POJ 1442·TREAP实现)