【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 = 30000 + 10;
const int SIZE = 250;//块状链表的大小 
const int M = 50000 + 5;
using namespace std;
struct TREAP{
       struct Node{
              int fix, size;
              int val;
              Node *ch[2];
       }mem[30000 + 10], *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 = 0;
            p->size = 1;
            p->ch[0] = p->ch[1] = 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 ^ 1];
            p->ch[d ^ 1] = t;
            t->size = 1;
            if (t->ch[0] != NULL) t->size += t->ch[0]->size;
            if (t->ch[1] != NULL) t->size += t->ch[1]->size; 
            t = p;
            t->size = 1;
            if (t->ch[0] != NULL) t->size += t->ch[0]->size;
            if (t->ch[1] != NULL) t->size += t->ch[1]->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 = 1;
            if (t->ch[0] != NULL) t->size += t->ch[0]->size;
            if (t->ch[1] != NULL) t->size += t->ch[1]->size; 
       }
       //在t的子树中找到第k小的值 
       int find(Node *t, int k){
           if (t->size == 1) return t->val;
           int l = 0;//t的左子树中有多少值 
           if (t->ch[0] != NULL) l += t->ch[0]->size;
           if (k == (l + 1)) return t->val;
           if (k <= l) return find(t->ch[0], k);
           else return find(t->ch[1], k - (l + 1));
       }
}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 = 0; 
     memset(have, 0, sizeof(have));
     scanf("%d%d", &m, &n);
     for (int i = 1; i <= m; i++) scanf("%d", &data[i]);
     for (int i = 1; i <= n; i++){
         int x;
         scanf("%d", &x);
         have[x]++;
     }
}
void work(){
     int pos = 0;//代表要获得的位置 
     for (int i = 1; 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 0;
}