【POJ1442】【Treap】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

【分析】

练下手而已，没什么。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <assert.h>
 #include <map>

 const int N =  + ;
 const int SIZE = ;//块状链表的大小
 const int M =  + ;
 using namespace std;
 struct TREAP{
        struct Node{
               int fix, size;
               int val;
               Node *ch[];
        }mem[ + ], *root;
        int tot;
        //大随机
        int BIG_RAND(){return (rand() * RAND_MAX + rand());}
        Node *NEW(){
             Node *p = &mem[tot++];
             p->fix = BIG_RAND();
             p->val = ;
             p->size = ;
             p->ch[] = p->ch[] = NULL;
             return p;
        }
        //将t的d节点换到t
        void rotate(Node *&t, int d){
             Node *p = t->ch[d];
             t->ch[d] = p->ch[d ^ ];
             p->ch[d ^ ] = t;
             t->size = ;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
             t = p;
             t->size = ;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
             return;
        }
        void insert(Node *&t, int val){
             //插入
             if (t == NULL){
                t = NEW();
                t->val = val;
                return;
             }
             //大的在右边,小的在左边
             int dir = (val >= t->val);
             insert(t->ch[dir], val);
             //维护最大堆的性质
             if (t->ch[dir]->fix > t->fix) rotate(t, dir);
             t->size = ;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
             if (t->ch[] != NULL) t->size += t->ch[]->size;
        }
        //在t的子树中找到第k小的值
        int find(Node *t, int k){
            if (t->size == ) return t->val;
            int l = ;//t的左子树中有多少值
            if (t->ch[] != NULL) l += t->ch[]->size;
            if (k == (l + )) return t->val;
            if (k <= l) return find(t->ch[], k);
            else return find(t->ch[], k - (l + ));
        }
 }treap;
 typedef long long ll;
 int have[N];//have为1则在这个地方GET
 int data[N], m, n;

 void init(){
      treap.root = NULL;
      treap.tot = ;
      memset(have, , sizeof(have));
      scanf("%d%d", &m, &n);
      for (int i = ; i <= m; i++) scanf("%d", &data[i]);
      for (int i = ; i <= n; i++){
          int x;
          scanf("%d", &x);
          have[x]++;
      }
 }
 void work(){
      int pos = ;//代表要获得的位置
      for (int i = ; i <= m; i++){
          treap.insert(treap.root, data[i]);
          //printf("%d", treap.root->size);
          while (have[i]){
             pos++;
             printf("%d\n", treap.find(treap.root, pos));
             have[i]--;
          }
      }

 }

 int main(){
     int T;
     #ifdef LOCAL
     freopen("data.txt", "r", stdin);
     freopen("out.txt", "w", stdout);
     #endif
     init();
     work();
     return ;
 }