NC50940 Running Median

题目链接

题目

题目描述

For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.

输入描述

The first line of input contains a single integer $P(1 \leq P \leq 1000)$ , which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer $M (1 \leq M \leq 9999)$ , giving the total number of signed integers to be processed. The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.

输出描述

For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.

示例1

输入

3 
1 9 
1 2 3 4 5 6 7 8 9 
2 9 
9 8 7 6 5 4 3 2 1 
3 23 
23 41 13 22 -3 24 -31 -11 -8 -7 
3 5 103 211 -311 -45 -67 -73 -81 -99 
-33 24 56

输出

1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3 
-7 -3

题解

知识点：优先队列。

用两个优先队列维护中位数左边和右边的序列，因为中位数一定在序列的中间位置，左边比他大，右边比他小且数量几乎一样（相差不超过 $1$ ）。因此左边维护大顶堆，右边维护小顶堆，每次存入数据后比较两序列的大小，那边比那边大超过 $1$ 就把队头转移直至平衡，就可以维护一个动态中位数了。

时间复杂度 $O(n \log n)$

空间复杂度 $O(n)$

代码

#include <bits/stdc++.h>
#define ll long long
 
using namespace std;
 
bool solve() {
    int p, m;
    cin >> p >> m;
    cout << p << ' ' << (m + 1) / 2 << '\n';
    priority_queue<int> pq1;
    priority_queue<int, vector<int>, greater<int>> pq2;
    int tmp;
    cin >> tmp;
    pq1.push(tmp);
    cout << pq1.top() << ' ';
    for (int i = 2;i <= m;i++) {
        cin >> tmp;
        if (tmp <= pq1.top()) pq1.push(tmp);
        else pq2.push(tmp);
        if (pq1.size() > 1 + pq2.size()) {
            pq2.push(pq1.top());
            pq1.pop();
        }
        else if (pq1.size() + 1 < pq2.size()) {///注意，不要size相减，会ULL溢出，-1变最大值
            pq1.push(pq2.top());
            pq2.pop();
        }
        if (i & 1) cout << (pq1.size() > pq2.size() ? pq1.top() : pq2.top()) << ' ';
        if (!(i % 20)) cout << '\n';
    }
    cout << '\n';
    return true;
}
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
        if (!solve()) cout << -1 << '\n';
    }
    return 0;
}