Black Box

链接:

  http://poj.org/problem?id=1442

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题目：

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

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题意：

  给你n个数，m个询问，然后第i个寻问Mi表示当盒子里有Mi个数的时候从小到大排名为i的数是多少。

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思路：

  先把n个数存进数组里，在询问为M的时候再将前M个数放入树堆，然后查询第i大就可以了。

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实现：

#include <iostream>#include <algorithm>#include <set>#include <string>#include <vector>#include <queue>#include <map>#include <stack>#include <list>#include <iomanip>#include <functional>#include <sstream>#include <cstdio>#include <cstring>#include <cmath>#include <cctype>#define read read()#define edl putchar('\n')#define clr(a, b) memset(a,b,sizeof a)inline int read {    int x = 0;    char c = getchar();    while (c < '0' || c > '9')c = getchar();    while (c >= '0' && c <= '9') {        x = x * 10 + c - '0';        c = getchar();    }    return x;}inline void write(int x) {    int y = 10, len = 1;    while (y <= x) {        y *= 10;        len++;    }    while (len--) {        y /= 10;        putchar(x / y + 48);        x %= y;    }}using namespace std;template <class T, class Compare = std::less<T> >class Treap {private:    struct treap {        int size, fix;        T key;        Compare cmp;        treap *ch[2];        treap(T key) {            size = 1;            fix = rand();            this->key = key;            ch[0] = ch[1] = NULL;        }        int compare(T x) const {            if (x == key) return -1;            return cmp(x ,key) ? 0 : 1;        }        void Maintain() {            size = 1;            if (ch[0] != NULL) size += ch[0]->size;            if (ch[1] != NULL) size += ch[1]->size;        }    }*root;    Compare cmp;    void Rotate(treap *&t, int d) {        treap *k = t->ch[d ^ 1];        t->ch[d ^ 1] = k->ch[d];        k->ch[d] = t;        t->Maintain();        k->Maintain();        t = k;    }    void Insert(treap *&t, T x) {        if (t == NULL) t = new treap(x);        else {            //int d = t->compare(x);            int d = cmp(x ,t->key) ? 0 : 1;            Insert(t->ch[d], x);            if (t->ch[d]->fix > t->fix) Rotate(t, d ^ 1);        }        t->Maintain();    }    void Delete(treap *&t, int x) {        int d = t->compare(x);        if (d == -1) {            treap *tmp = t;            if (t->ch[0] == NULL) {                t = t->ch[1];                delete tmp;                tmp = NULL;            } else if (t->ch[1] == NULL) {                t = t->ch[0];                delete tmp;                tmp = NULL;            } else {                int k = t->ch[0]->fix > t->ch[1]->fix ? 1 : 0;                Rotate(t, k);                Delete(t->ch[k], x);            }        } else Delete(t->ch[d], x);        if (t != NULL) t->Maintain();    }    bool Find(treap *t, int x) {        while (t != NULL) {            int d = t->compare(x);            if (d == -1) return true;            t = t->ch[d];        }        return false;    }    T Kth(treap *t, int k) {        if (t == NULL || k <= 0 || k > t->size) return -1;        if (t->ch[0] == NULL) {            if (k == 1) return t->key;            return Kth(t->ch[1], k - 1);        }        if (t->ch[0]->size >= k) return Kth(t->ch[0], k);        if (t->ch[0]->size + 1 == k) return t->key;        return Kth(t->ch[1], k - 1 - t->ch[0]->size);    }    int Rank(treap *t, int x) {        int r;        if (t->ch[0] == NULL) r = 0;        else r = t->ch[0]->size;        if (x == t->key) return r + 1;        if (x < t->key) return Rank(t->ch[0], x);        return r + 1 + Rank(t->ch[1], x);    }    void Deletetreap(treap *&t) {        if (t == NULL) return;        if (t->ch[0] != NULL) Deletetreap(t->ch[0]);        if (t->ch[1] != NULL) Deletetreap(t->ch[1]);        delete t;        t = NULL;    }    void Print(treap *t) {        if (t == NULL) return;        Print(t->ch[0]);        cout << t->key << ' ';        Print(t->ch[1]);    }public:    Treap() {        root = NULL;    }    ~Treap() {        Deletetreap(root);    }    void insert(T x) {        Insert(root, x);    }    void clear() {        Deletetreap(root);    }    void print() {        Print(root);    }    T kth(int x) {        return Kth(root, x);    }};int val[int(1e6) + 7];struct A {    int x;    A(int x):x(x) {}    A() {}    bool operator < (const A &tmp) const {        return x < tmp.x;    }};int main() {    ios_base::sync_with_stdio(false);    cin.tie(0);#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif    int n, x, m;    Treap<A> a;    while (a.clear(), ~scanf("%d%d", &n, &m)) {        for (int i = 1; i <= n; i++)            scanf("%d", &val[i]);        int index = 1;        for (int i = 1; i <= m; i++) {            scanf("%d", &x);            for (int j = index; j <= x; j++)                a.insert(A{val[j]});            index = x + 1;            printf("%d\n", a.kth(i).x);        }    }    return 0;}