poj1442--Black Box--优先队列维护堆&&multiset维护堆

Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6002 Accepted: 2420

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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

Source

Northeastern Europe 1996

这道题学到两个东西：

1.怎样通过结构体定义使用priority_queue

2.怎样在结构体里面重载<运算符从而省去mycompare函数

3.Priority_queue确实比multiset常优，但是写起来烦。

4.C++中有太多东西是左闭右开的了，比如mulitiset  a中的a.end()，返回的并不是a中最后一个元素的指针。

题目有点长，题意还好。。。不算抽。

题意就是，给你N个查询ai，第i个查询加入ai个数后，已排好序的ai个数中第i个数的大小。

做法就是，用两个堆来维护，一个最小堆，一个最大堆。

在最大堆中放i-1个数，在最小堆中放第i个到第N个元素，

维护一点，最小堆的顶部（最小元素）大于最大堆顶部（最大元素），

那么最小堆的顶部即为所求的答案。

首先用单调队列写了一个。

以下是代码+注释，

12105497lovellp1442Accepted672K157MSC++1394B2013-09-13 10:54:18

#include <iostream>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxnum=3e4+100;struct max_heap{    int x;    bool operator<(const max_heap&a) const //第一次写重载运算符，感觉还蛮好ORZ,可以少写一个my_compare函数了    {        return x<a.x;    }};struct min_heap{    int x;    bool operator<(const min_heap&a) const //这个运算符是相反定义的，因而通过单调队列它维护了一个最小堆。    {        return x>a.x;    }};int a[maxnum];priority_queue<max_heap> max_h;priority_queue<min_heap> min_h;max_heap max_t;//这个东东神奇的功效接下来就会看到min_heap min_t;//这个东东神奇的功效接下来就会看到int main(){    //freopen("input.txt","r",stdin);    int m,n;    cin>>m>>n;    for(int i=1;i<=m;++i)        scanf("%d",&a[i]);//先读入所有数据，再慢慢进行处理    int num=1;    for(int i=1;i<=n;++i)    {        int t;        scanf("%d",&t);        while(num<=t) //这里很精妙，只有当所给查询位置大于已加入元素个数的时候才进行添加        {            if(!max_h.empty()&&max_h.top().x>a[num])//只有当最大堆非空才进行比较！这点很重要！            {                int temp=max_h.top().x;                max_h.pop();                max_t.x=a[num];                max_h.push(max_t);                min_t.x=temp;                min_h.push(min_t);            }            else            {                min_t.x=a[num];                min_h.push(min_t);            }            ++num;        }        printf("%d\n",min_h.top().x);//输出        //接下来更是精妙的所在，即每过一个查询，都要维护最大堆的个数+1！        int temp=min_h.top().x;        max_t.x=temp;        min_h.pop();        max_h.push(max_t);    }    return 0;}

后来想想multiset好像也能实现这个功能，又用multiset写了一个，

即用Multiset维护两个堆，

大体思想不变，

但是如学长所说～～～Multiset作为非线性结构，的确比priotity_queue慢了常数级～～～

12106171lovellp1442Accepted1708K688MSC++1179B2013-09-13 15:15:40

以下是代码+注释：

#include <iostream>#include <cstdio>#include <set>using namespace std;const int maxnum=3e4+100;int hehe[maxnum];multiset<int> a;multiset<int> b;int main(){    //freopen("input.txt","r",stdin);    int n,q;    scanf("%d%d",&n,&q);    for(int i=1;i<=n;++i)        scanf("%d",&hehe[i]);    int num=1;    for(int i=1;i<=q;++i)    {        int t;        scanf("%d",&t);        //update data.        while(t>=num)        {            if(!b.empty())            {                multiset<int>::iterator b_it=b.end();                --b_it;   //这里一定要注意！调了好久的bug。要切记，C++有太多的东西是左闭右开的了!                if(hehe[num]<*b_it)                {                    int temp=*b_it;                    b.erase(b_it);                    b.insert(hehe[num]);                    a.insert(temp);                }                else a.insert(hehe[num]);            }            else            {                a.insert(hehe[num]);            }            ++num;        }        //cout data        multiset<int>::iterator a_it=a.begin();        printf("%d\n",*a_it);        //update multiset b        int temp=*a_it;        a.erase(a_it);        b.insert(temp);    }    return 0;}