TZOJ 2943 Running Median(动态中位数)

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

题意

n个数输出每输入奇数个时的中位数。

题解

动态中位数经典题。

由于输出奇数个数，相当于中位数只有1个。不会有除2的情况。

维护两个优先队列，1个从大到小pq1，1个从小到大pq2。

需要做到，pq1.top就是中位数，那么相当于pq1的个数等于pq2的个数或者多1个，并且pq1中的数<=pq2中的数。

那么显然对于一个数x，如果目前的中位数pq1.top>x，放到pq1，否则放到pq2。

放到pq1后，会出现pq1的大小比pq2大2，相当于pq1.top不是中位数了，中位数是pq1.pop后的pq1.top，当然pq1.pop后要把数放到pq2中。

放到pq2后，会出现pq2的大小比pq1大，相当于中位数在pq2.top，由于维护的是pq2的个数要小于pq1，那么pq2.pop后放到pq1中，pq1.top还是中位数。

<扩展 如果偶数也要输出怎么办?>

<显然就是经过维护后的(pq1.top+pq2.top)/2>

代码

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t,x,ca,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&ca,&n);
        printf("%d %d\n",ca,n/2+1);
        priority_queue< int,vector<int>,less<int> > pq1;
        priority_queue< int,vector<int>,greater<int> >pq2;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(pq1.empty()||pq1.top()>x)pq1.push(x);
            else pq2.push(x);
            if(pq1.size()>pq2.size()+1)
            {
                pq2.push(pq1.top());
                pq1.pop();
            }
            else if(pq1.size()<pq2.size())
            {
                pq1.push(pq2.top());
                pq2.pop();
            }
            if(i&1)
                printf("%d%c",pq1.top(),i==n?'\n':((i+1)%20?' ':'\n'));
        }
    }
    return 0;
}