题目：
Description
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.
Input
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.
Output
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.
Sample Input
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
Sample Output
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
题目大意：
依次给出n个数字，每输入一个奇数则输出即时中位数。（共有(n+1)/2个）
解题思路：
对顶堆，即用两个堆，一个大根堆一个小根堆。每次比较两个堆的堆顶，如果不相等就交换堆顶，否则堆顶即为所要求的中位数。
然后还是一如既往的使用C++的骚库
Accepted code：
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<cctype>
inline void read(int &x)
{
char c=getchar(); int f=1;x=0;
while(!isdigit(c)) {if (c=='-') f=-1;c=getchar();}
while(isdigit(c)) {x=x*10+c-48;c=getchar();}
x*=f; return;
}
std::priority_queue <int> lq,sq;
//lq大根堆，sq小根堆
int main()
{
int n,t,x,a,b,cnt;
read(t);
while(t--)
{
read(cnt); read(n);
while(lq.size()) lq.pop();
while(sq.size()) sq.pop();
printf("%d %d\n",cnt,(n+1)/2);
for(int i=1;i<=n;i++)
{
read(x);
lq.push(x),sq.push(-x);
if(i%2==0) continue;
while(lq.top()!=-sq.top())
{
a=lq.top();lq.pop();
b=-sq.top();sq.pop();
sq.push(-a);lq.push(b);
}
printf("%d ",lq.top());
if(((i+1)/2)%10 == 0) puts("");
else if((n%2==1&&i==n)||(n%2==0&&i==n-1))
puts("");
}
}
return 0;
}