Running Median

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

The first line of input contains a single integer P,(1 ≤ P ≤ 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 ≤ M ≤ 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

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.

样例输入

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

样例输出

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

code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        int id,n;
        scanf("%d%d",&id,&n);
        printf("%d %dn",id,(n+1)/2);
        priority_queue<int> l;
        priority_queue<int,vector<int>,greater<int>> r;
        for(int i = 1; i <= n; ++i){
            int x;
            scanf("%d",&x);
            if(l.size()>r.size()&&x<l.top())
                r.push(l.top()),l.pop(),l.push(x);
            else r.push(x);

            while(l.size()<r.size()) l.push(r.top()),r.pop();
            int j=(i+1)/2;
            if(i&1) printf("%d%s",l.top(),(i<n&&j%10!=0)?" ":"n");
        }
    }
    return 0;
}