poj1442 Black Box 堆维护

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前k小放在一个大顶堆里,每要插入一个数,维护这个大顶堆存前k - 1小,第k小就是小顶堆的top

小顶堆存k+缓存

附:今天福建省第七届大学生程序设计竞赛重现的C题很简单,就是求一个数学期望,由于n个数互不相同,所以任取两个数,都有一个较大的数,不会出现平局的情况,所以胖哥的得分期望就是n*0.5,就是填空题,不用管n张扑克牌的值。

Black Box
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 13119 Accepted: 5354

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

#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <queue>#include <functional> //小顶堆是用得到greaterusing namespace std;priority_queue<int, vector<int>, less<int> > maxHeap; //大顶堆priority_queue<int, vector<int>, greater<int> > minHeap;const int maxn = 30010;int a[maxn], b[maxn];int main(){int n, m;while (~scanf("%d%d", &n, &m)) {for (int i = 1; i <= n; ++i)scanf("%d", a + i);for (int i = 1; i <= m; ++i)scanf("%d", b + i);int j = 1, k = 0;//前k小放在一个大顶堆里,每要插入一个数,维护这个大顶堆存前k - 1小,第k小就是小顶堆的top//小顶堆存k+缓存for (int i = 1; i <= n; ++i) {if (maxHeap.size() < k) {if (!minHeap.empty() && minHeap.top() < a[i])maxHeap.push(minHeap.top()), minHeap.pop(), minHeap.push(a[i]);else maxHeap.push(a[i]);}else {if (!maxHeap.empty() && maxHeap.top() > a[i])minHeap.push(maxHeap.top()), maxHeap.pop(), maxHeap.push(a[i]);else minHeap.push(a[i]);}while (j <= m && i == b[j]) {while (maxHeap.size() < k) {int tmp = minHeap.top(); minHeap.pop(); //只要后台输入合法,放心top,popmaxHeap.push(tmp);}printf("%d\n", minHeap.top());++k;++j;}}while (!maxHeap.empty()) maxHeap.pop();while (!minHeap.empty()) minHeap.pop();}return 0;}



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