POJ 3784 Running Median

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：依次添加 n 个数，要求第奇数次添加时，输出当前数列中的中位数。
然而，POJ的输入输出总是十分奇葩。。。
维护两个堆，一个大根堆，一个小根堆 显然应该让大根堆存较小的数，小根堆存较大的数 如果维护好了这两个堆，令当前的中位数就是小根堆的堆顶 每次新加入一个数，如果这个数比当前的中位数大，就存入小根堆， 否则存入大根堆 限制小根堆中元素的个数比大根堆多 1 或相等 如果个数相差太多，则将元素多的堆删掉堆顶插入到元素少的堆中 此时就可以保证当前的中位数就是小根堆的堆顶 时间复杂度 O(nlogn)

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
priority_queue<int>bg;
priority_queue< int,vector<int>,greater<int> >sm;
int main()
{
    int q,x;
    scanf("%d",&q);
    while(q--)
    {
        int d,n,t=0;
        scanf("%d%d",&d,&n);
        printf("%d %d\n",d,(n+1)/2);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(sm.empty()||x>sm.top())
                sm.push(x);
            else
                bg.push(x);
            if(i%2==1)
            {
                while(sm.size()>bg.size()+1)
                {
                    x=sm.top();
                    sm.pop();
                    bg.push(x);
                }
                while(sm.size()<bg.size()+1)
                {
                    x=bg.top();
                    bg.pop();
                    sm.push(x);
                }
                t++;
                if(t%10!=0)
                    printf("%d ",sm.top());
                else
                    printf("%d\n",sm.top());
            }
        }
        if(((n+1)/2)%10!=0)
            printf("\n");
        while(!sm.empty())
            sm.pop();
        while(!bg.empty())
            bg.pop();
    }
    return 0;
}