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, 8It 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
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 #include2 #include 3 #include 4 #include 5 #include 6 #include 7 #include 8 #include 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 }