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: 10000K | |
|---|---|---|
| Total 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 4
3 1 -4 2 8 -1000 2
1 2 6 6
Sample Output
3
3
1
2
Source
\
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <queue>
#include <functional> //小顶堆是用得到greater
using 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,pop
maxHeap.push(tmp);
}
printf("%d\n", minHeap.top());
++k;++j;
}
}
while (!maxHeap.empty()) maxHeap.pop();
while (!minHeap.empty()) minHeap.pop();
}
return 0;
}
\
\
