Black Box+POJ+treap树模板
2014-09-03 14:22
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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
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
Sample Output
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Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 7437 | Accepted: 3051 |
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; }
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