题目正文
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.
翻译
对于这个问题，您将编写一个以32位有符号整数序列读取的程序。读取每个奇数索引值后，输出到目前为止接收到的元素的中值（中间值）。
输入
输入的第一行包含一个整数P，（1≤ P≤ 1000），这是后面的数据集数。每个数据集的第一行包含数据集编号，后跟一个空格，后跟一个奇数十进制整数M（1）≤ M≤ 9999），给出要处理的有符号整数的总数。
数据集中的其余行由值组成，每行10个，用单个空格分隔。
数据集中的最后一行可能包含少于10个值。
输出
对于每个数据集，输出的第一行包含数据集编号、单个空格和输出的中位数（应为输入值数量的一半加一）。输出中位数将位于以下行上，每行10个，由一个空格分隔。最后一行可以有少于10个元素，但至少有1个元素。输出中不应有空行。
代码
代码：
#include<stdio.h>
#include<iostream>
#include<set>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int x,n,a;
scanf("%d%d",&x,&n);
priority_queue<int> p1;
priority_queue<int,vector<int>,greater<int>> p2;
vector<int> ans;
for(int i=1;i<=n;i++)
{
scanf("%d",&a);
if(p1.empty())
p1.push(a);
else
{
if(a>p1.top())
p2.push(a);
else
p1.push(a);
}
while(p1.size()<p2.size())
{
p1.push(p2.top());
p2.pop();
}
while(p2.size()<p1.size()-1)
{
p2.push(p1.top());
p1.pop();
}
if(i&1)
ans.push_back(p1.top());
}
printf("%d %d\n",x,ans.size());
for(int i=1;i<=ans.size();i++)
{
if(i%10!=1)
printf(" ");
printf("%d",ans[i-1]);
if(i%10==0)
printf("\n");
}
printf("\n");
}
}
总结
对顶推解法，找出中位数