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 $P(1 \le P \le 1000)$, 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 $M(1 \le M \le 9999)$, 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.

输入样例:

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

输出样例:

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

题目分析

Heap

代码实现

#pragma GCC optimize O(2)
#include <iostream>
#include <queue>
using namespace std;
const int N = 1e5 + 10;

int p, idx, n, a[N];
int main () {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> p;
    while(p--) {
        cin >> idx >> n;
        int cnt = 1;
        priority_queue <int> h;
        priority_queue <int, vector <int>, greater <int>> H;
        cout << idx << ' ' << ((n + 1) >> 1) << '\n';
        cin >> a[1];
        h.push(a[1]);
        cout << a[1] << ' ';
        for(int i = 2; i <= n; ++i) {
            ++cnt;
            if(cnt
            cin >> a[i];
            if(h.top() < a[i]) H.push(a[i]);
            else h.push(a[i]);
            if(i & 1) {
                while(h.size() > (i + 1) >> 1) {
                    H.push(h.top());
                    h.pop();
                }
                while(H.size() > i >> 1) {
                    h.push(H.top());
                    H.pop();
                }
                cout << h.top() << ' ';
            }
        }
        cout << '\n';
    }
    return 0;
}
分类: Heap

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