第k大数维护，我推荐Treap。。谁用谁知道。。。。
Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 5999 | Accepted: 2417 |
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.
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
|
#include <iostream>
#include <cstdio> #include <cstdlib> #include <cstring> using namespace std; const int maxNode=444444; const int INF=0x3f3f3f3f; int M,N,II,cc; int arr[33000]; int u[33000]; struct Treap { int root,treapCnt,key[maxNode],priority[maxNode], childs[maxNode][2],cnt[maxNode],ssize[maxNode]; Treap() { root=0; treapCnt=1; priority[0]=INF; ssize[0]=0; } void update(int x) { ssize[x]=ssize[childs[x][0]]+cnt[x]+ssize[childs[x][1]]; } void rotate(int &x,int t) { int y=childs[x][t]; childs[x][t]=childs[y][1-t]; childs[y][1-t]=x; update(x); update(y); x=y; } void _insert(int &x,int k) { if(x) { if(key[x]==k) { cnt[x]++; } else { int t=key[x]<k; _insert(childs[x][t],k); if(priority[childs[x][t]]<priority[x]) { rotate(x,t); } } } else { x=treapCnt++; key[x]=k; cnt[x]=1; priority[x]=rand(); childs[x][0]=childs[x][1]=0; } update(x); } int _getKth(int &x,int k) { if(k<=ssize[childs[x][0]]) { return _getKth(childs[x][0],k); } k-=ssize[childs[x][0]]+cnt[x]; if(k<=0) { return key[x]; } return _getKth(childs[x][1],k); } void Insert(int k) { _insert(root,k); } int GetKth(int k) { return _getKth(root,k); } }T; int main() { scanf("%d%d",&M,&N); II=0;cc=1; for(int i=1;i<=M;i++) { scanf("%d",arr+i); } for(int i=1;i<=N;i++) { scanf("%d",u+i); } for(int i=1;i<=M;i++) { T.Insert(arr[i]); while(i==u[cc]) { cc++;II++; printf("%d ",T.GetKth(II)); } } return 0; } |